🧠:[<think>Okay, so I need to find the minimum distance from a point P that moves along the curve ρ² cosθ - 2ρ = 0 to the fixed point Q which has polar coordinates (1, π/3). Hmm, let me start by understanding the problem step by step.First, let's recall that in polar coordinates, a point is represented as (ρ, θ), where ρ is the distance from the origin (or pole) and θ is the angle measured from the positive x-axis. So, point Q is at a distance of 1 from the origin and at an angle of π/3 radians, which is 60 degrees. That places Q in the first quadrant, right? Since π/3 is 60 degrees, so x = 1 * cos(π/3) = 0.5, y = 1 * sin(π/3) ≈ 0.866. So, Q is at (0.5, √3/2) in Cartesian coordinates.Now, the curve that point P moves on is given by the equation ρ² cosθ - 2ρ = 0. Let me try to simplify this equation. Let's factor out ρ: ρ(ρ cosθ - 2) = 0. So, either ρ = 0 or ρ cosθ - 2 = 0. But ρ = 0 is just the origin, so the actual curve is ρ cosθ - 2 = 0, which simplifies to ρ = 2 / cosθ. That's valid as long as cosθ ≠ 0, which would be θ ≠ π/2 or 3π/2. So, this is a polar equation, and I can try to convert it to Cartesian coordinates to see what kind of curve it is.In Cartesian coordinates, ρ = 2 / cosθ. Since ρ = √(x² + y²) and cosθ = x / ρ. Therefore, substituting cosθ into the equation gives:ρ = 2 / (x / ρ) => ρ = 2ρ / x => Multiply both sides by x: ρx = 2ρ. If ρ ≠ 0, we can divide both sides by ρ: x = 2. So, this is the vertical line x = 2 in Cartesian coordinates. Wait, that's interesting! So, the curve that P moves on is simply the vertical line x = 2. So, in polar coordinates, ρ = 2 / cosθ is equivalent to x = 2 in Cartesian.Therefore, point P is any point on the vertical line x = 2. So, the problem reduces to finding the minimum distance from any point on the line x = 2 to the point Q (0.5, √3/2). But wait, in Cartesian coordinates, the distance from a point (x, y) to (0.5, √3/2) is √[(x - 0.5)² + (y - √3/2)²]. Since all points P are on x = 2, their coordinates are (2, y) where y can be any real number. So, the distance squared would be (2 - 0.5)² + (y - √3/2)² = (1.5)² + (y - √3/2)². To minimize this distance, we can minimize the square of the distance, which is easier.The minimum occurs when the second term is zero, i.e., when y = √3/2. Therefore, the closest point on x = 2 to Q is (2, √3/2), and the minimum distance is √[(1.5)² + 0] = 1.5 = 3/2. So, the minimum distance is 3/2.But wait a second! The problem is stated in polar coordinates, so maybe I need to verify this using polar coordinates rather than converting to Cartesian. Let me check.In polar coordinates, the distance between two points (ρ1, θ1) and (ρ2, θ2) is given by the formula:d = √[ρ1² + ρ2² - 2ρ1ρ2 cos(θ1 - θ2)]So, here, point Q is (1, π/3) and point P is (ρ, θ) on the curve ρ² cosθ - 2ρ = 0, which we converted to ρ = 2 / cosθ. So, substituting ρ = 2 / cosθ into the distance formula:d = √[(2 / cosθ)² + 1² - 2*(2 / cosθ)*1*cos(θ - π/3)]Let me compute this step by step.First, expand (2 / cosθ)²: 4 / cos²θThen, the second term is 1² = 1.Third term: -2*(2 / cosθ)*1*cos(θ - π/3) = -4 / cosθ * cos(θ - π/3)Therefore, d² = 4 / cos²θ + 1 - 4 / cosθ * cos(θ - π/3)Our goal is to minimize d² with respect to θ. Let's denote f(θ) = 4 / cos²θ + 1 - 4 / cosθ * cos(θ - π/3). We need to find the θ that minimizes f(θ).This seems more complicated, but maybe we can simplify it. Let me see.First, note that 1 / cosθ is secθ, so f(θ) = 4 sec²θ + 1 - 4 secθ cos(θ - π/3)Let me express cos(θ - π/3) using the cosine subtraction formula:cos(θ - π/3) = cosθ cosπ/3 + sinθ sinπ/3 = (1/2)cosθ + (√3/2) sinθTherefore, substituting back into f(θ):f(θ) = 4 sec²θ + 1 - 4 secθ [ (1/2)cosθ + (√3/2) sinθ ]Simplify the term inside the brackets:(1/2)cosθ + (√3/2) sinθ. Since secθ = 1 / cosθ, then secθ * cosθ = 1, and secθ * sinθ = tanθ.Therefore, 4 secθ [ (1/2)cosθ + (√3/2) sinθ ] = 4 [ (1/2) + (√3/2) tanθ ] = 4*(1/2) + 4*(√3/2) tanθ = 2 + 2√3 tanθTherefore, f(θ) simplifies to:4 sec²θ + 1 - [2 + 2√3 tanθ] = 4 sec²θ + 1 - 2 - 2√3 tanθ = 4 sec²θ -1 - 2√3 tanθSo, f(θ) = 4 sec²θ - 2√3 tanθ -1Now, let's express everything in terms of tanθ. Since sec²θ = 1 + tan²θ, so:f(θ) = 4(1 + tan²θ) - 2√3 tanθ -1 = 4 + 4 tan²θ - 2√3 tanθ -1 = 3 + 4 tan²θ - 2√3 tanθLet me set u = tanθ. Then f(θ) becomes:f(u) = 4u² - 2√3 u + 3This is a quadratic in u. To find the minimum of f(u), since the coefficient of u² is positive (4), the parabola opens upwards, so the minimum is at the vertex.The vertex of a quadratic ax² + bx + c is at u = -b/(2a). Here, a = 4, b = -2√3. Therefore:u = -(-2√3)/(2*4) = (2√3)/8 = √3/4Therefore, the minimum occurs when tanθ = √3/4So, θ = arctan(√3 / 4)Let me compute that. But perhaps we don't need the exact angle, just the value of f(u) at the minimum.Substituting u = √3/4 into f(u):f(u) = 4*( (√3/4)^2 ) - 2√3*(√3/4) + 3Calculate each term:First term: 4*(3/16) = 12/16 = 3/4Second term: 2√3*(√3/4) = 2*(3)/4 = 6/4 = 3/2Third term: 3So, f(u) = 3/4 - 3/2 + 3 = (3/4 - 6/4 + 12/4) = (3 -6 +12)/4 = 9/4Therefore, the minimal d² is 9/4, so the minimal distance d is 3/2, which is 1.5, same as before. So, this confirms the previous result. Therefore, the minimal distance is 3/2.But let me cross-verify once again. When I converted the polar equation to Cartesian, I found that it's x = 2. Then, the distance from x=2 to (0.5, √3/2) is indeed 2 - 0.5 = 1.5 horizontally, and since the point on x=2 with the same y-coordinate as Q is (2, √3/2), so vertical distance is zero, so the minimal distance is 1.5. So both methods agree. Therefore, the answer should be 3/2.But just to make sure, let's check with calculus. Suppose we didn't convert to Cartesian and stuck with polar coordinates. Let's see.The curve is ρ = 2 / cosθ. The distance from any point P(ρ, θ) to Q(1, π/3) is given by the formula:d = sqrt(ρ² + 1² - 2*ρ*1*cos(θ - π/3))We need to minimize d, which is equivalent to minimizing d² = ρ² + 1 - 2ρ cos(θ - π/3)But since ρ = 2 / cosθ, substitute that in:d² = (4 / cos²θ) + 1 - 2*(2 / cosθ)*cos(θ - π/3)Which is exactly the same expression as before. Then, as we saw earlier, simplifying leads to d² = 9/4, so d = 3/2. So, this confirms again.Alternatively, using calculus, treating θ as the variable, take derivative of f(θ) = 4 sec²θ -2√3 tanθ -1, set derivative to zero.But maybe let's try this approach.First, express f(θ) = 4 sec²θ - 2√3 tanθ -1Compute derivative f’(θ):d/dθ [4 sec²θ] = 4 * 2 secθ * secθ tanθ = 8 sec²θ tanθd/dθ [ -2√3 tanθ ] = -2√3 sec²θd/dθ [ -1 ] = 0Therefore, f’(θ) = 8 sec²θ tanθ - 2√3 sec²θFactor out 2 sec²θ:f’(θ) = 2 sec²θ (4 tanθ - √3 )Set f’(θ) = 0:2 sec²θ (4 tanθ - √3 ) = 0Since sec²θ is never zero, the critical points occur when 4 tanθ - √3 = 0 => tanθ = √3 /4, which is the same result as before. Thus, θ = arctan(√3 /4). So, same conclusion.Therefore, the minimal distance is indeed 3/2.But let me just think again if there could be another critical point or if there's a mistake in the process. For example, in polar coordinates, sometimes θ is restricted to [0, 2π), but since tanθ is periodic with period π, arctan(√3 /4) will give a value in (-π/2, π/2), but since θ in polar coordinates can be adjusted by adding π, but since ρ is positive, and our original equation ρ = 2 / cosθ requires that cosθ is positive (since ρ must be positive), so θ must be in (-π/2, π/2). Therefore, arctan(√3 /4) is in the first or fourth quadrant. However, since √3 /4 is positive, θ is in the first quadrant. So, θ is between 0 and π/2. Therefore, the minimal distance occurs at θ = arctan(√3 /4). But since we converted to Cartesian and found the minimal distance is 3/2, which is achieved at (2, √3/2), let's check what θ that corresponds to.The point (2, √3/2) in Cartesian is (x=2, y=√3/2). Converting to polar coordinates:ρ = sqrt(2² + (√3/2)^2) = sqrt(4 + 3/4) = sqrt(19/4) = sqrt(19)/2 ≈ 2.179θ = arctan(y/x) = arctan( (√3/2)/2 ) = arctan(√3 /4 ) ≈ arctan(0.433) ≈ 0.410 radians, which is about 23.5 degrees. But in our previous calculation, we found that θ = arctan(√3 /4). So, this matches. Therefore, the minimal distance occurs at θ = arctan(√3 /4), and the minimal distance is 3/2.But let me check if the point (2, √3/2) is indeed on the curve x=2. Yes, since x=2, so that point is on the line. Therefore, the minimal distance is 3/2.Alternatively, if we consider the line x=2 and the point (0.5, √3/2), the minimal distance is the horizontal distance between x=2 and x=0.5, which is 1.5, since the vertical coordinate can be matched. Therefore, the minimal distance is indeed 3/2.Therefore, all methods converge to the same answer. Thus, the minimum distance is 3/2.Final AnswerThe minimum distance from point ( P ) to point ( Q ) is boxed{dfrac{3}{2}}.</think>Given that point ( P ) moves on the curve ( rho^2 cos theta - 2rho = 0 ) in the polar coordinate system, we need to find the minimum distance from point ( P ) to point ( Q(1, frac{pi}{3}) ).First, we simplify the given curve equation:[rho^2 cos theta - 2rho = 0 implies rho(rho cos theta - 2) = 0]Since (rho neq 0), we have (rho cos theta - 2 = 0 implies rho = frac{2}{cos theta}). This is the polar equation of the vertical line ( x = 2 ) in Cartesian coordinates.Next, we convert the polar coordinates of point ( Q ) to Cartesian coordinates:[Qleft(1, frac{pi}{3}right) implies left(0.5, frac{sqrt{3}}{2}right)]The minimum distance from a point on the line ( x = 2 ) to the point ( left(0.5, frac{sqrt{3}}{2}right) ) is the horizontal distance between ( x = 2 ) and ( x = 0.5 ), which is ( 2 - 0.5 = 1.5 ). This distance is achieved when the vertical coordinate of ( P ) matches that of ( Q ), i.e., ( y = frac{sqrt{3}}{2} ).To confirm, we use the distance formula in polar coordinates. The distance ( d ) between points ( (rho, theta) ) and ( (1, frac{pi}{3}) ) is:[d = sqrt{rho^2 + 1 - 2rho cos(theta - frac{pi}{3})}]Substituting (rho = frac{2}{cos theta}) into the distance formula and simplifying, we find the minimum distance occurs when (theta = arctanleft(frac{sqrt{3}}{4}right)), leading to the minimum distance squared of (frac{9}{4}), hence the minimum distance is (frac{3}{2}).Thus, the minimum distance from point ( P ) to point ( Q ) is boxed{dfrac{3}{2}}.,<think>Okay, so I need to find the minimum distance from a point P moving on the curve ρ² cosθ - 2ρ = 0 to the fixed point Q(1, π/3) in polar coordinates. Hmm, let me start by understanding the problem step by step.First, let me recall that in polar coordinates, any point is represented as (ρ, θ), where ρ is the radius (distance from the origin) and θ is the angle made with the positive x-axis. The given curve for point P is ρ² cosθ - 2ρ = 0. Let me rewrite that equation to maybe make it more understandable. Let's see:ρ² cosθ - 2ρ = 0.I can factor out a ρ from the first two terms:ρ(ρ cosθ - 2) = 0.So, this equation is satisfied when either ρ = 0 or ρ cosθ - 2 = 0. However, ρ = 0 would just be the origin, but since P is moving along the entire curve, the main part of the curve is given by ρ cosθ - 2 = 0, which simplifies to ρ = 2 / cosθ. Ah, right, so ρ = 2 secθ. That makes sense. So, the curve is ρ = 2 secθ, which is a straight line in polar coordinates. Wait, because ρ = 2 secθ is equivalent to ρ cosθ = 2, which in Cartesian coordinates is x = 2. So, the curve is the vertical line x = 2. Got it. So, point P moves along the vertical line x = 2. And we need to find the minimum distance from this line to the point Q, which is given in polar coordinates as (1, π/3). Let me confirm the Cartesian coordinates for Q. Since polar coordinates (r, θ) convert to Cartesian as (r cosθ, r sinθ). So for Q(1, π/3):x = 1 * cos(π/3) = 1 * 0.5 = 0.5y = 1 * sin(π/3) = 1 * (√3 / 2) ≈ 0.866So Q is at (0.5, √3/2). And the curve P is moving on is x = 2. So, in Cartesian terms, we need the minimum distance from the line x=2 to the point (0.5, √3/2). Wait a second, is that right? If so, then maybe there's a simpler way here.But let me check. If the curve is x = 2, then the distance from any point (2, y) on this line to Q(0.5, √3/2) is sqrt[(2 - 0.5)^2 + (y - √3/2)^2]. To find the minimum distance, we can minimize this expression. Since sqrt is a monotonic function, the minimum occurs at the same point where the squared distance is minimized. So, let's compute the squared distance:(2 - 0.5)^2 + (y - √3/2)^2 = (1.5)^2 + (y - √3/2)^2 = 2.25 + (y - √3/2)^2.To minimize this, we take the derivative with respect to y and set it to zero. The derivative is 2(y - √3/2). Setting that to zero gives y = √3/2. Therefore, the minimum distance occurs when y = √3/2, so the closest point on the line x=2 to Q is (2, √3/2). Then the distance is sqrt[(2 - 0.5)^2 + (√3/2 - √3/2)^2] = sqrt[1.5^2 + 0] = 1.5 = 3/2. So the minimum distance is 3/2.But wait, this seems straightforward, but the problem is given in polar coordinates. Maybe there's a catch here? Let me verify if my conversion of the curve and the point is correct.Original curve: ρ² cosθ - 2ρ = 0. As I did earlier, factor out ρ: ρ(ρ cosθ - 2) = 0. So ρ = 0 or ρ = 2 / cosθ. ρ = 2 / cosθ is the same as ρ = 2 secθ, which in Cartesian is x = 2, as ρ cosθ = x. So yes, that's correct.Point Q is (1, π/3) in polar, which is indeed (0.5, √3/2) in Cartesian. So converting everything to Cartesian makes sense here.But maybe the problem expects me to solve it purely in polar coordinates? Let me try that approach as well to confirm.In polar coordinates, the distance between two points (ρ, θ) and (a, φ) is given by sqrt[ρ² + a² - 2ρ a cos(θ - φ)]. So here, point P is (ρ, θ) on the curve ρ = 2 secθ, and point Q is (1, π/3). So the distance squared between P and Q is:ρ² + 1² - 2 * ρ * 1 * cos(θ - π/3).But since ρ = 2 secθ, we can substitute that into the distance formula:Distance squared = (2 secθ)^2 + 1 - 2 * 2 secθ * cos(θ - π/3)Simplify:= 4 sec²θ + 1 - 4 secθ cos(θ - π/3)Hmm, this seems more complicated. Let's see if we can simplify this expression.First, note that secθ is 1/cosθ, so:4 sec²θ = 4 / cos²θSimilarly, 4 secθ cos(θ - π/3) = 4 * (1 / cosθ) * cos(θ - π/3)So, Distance squared = 4 / cos²θ + 1 - 4 cos(θ - π/3) / cosθLet me see if I can simplify the term 4 cos(θ - π/3) / cosθ.Using the cosine subtraction formula: cos(θ - π/3) = cosθ cosπ/3 + sinθ sinπ/3 = cosθ * 0.5 + sinθ * (√3 / 2)Therefore:4 [0.5 cosθ + (√3 / 2) sinθ] / cosθ = 4 [0.5 + (√3 / 2) tanθ] = 4 * 0.5 + 4 * (√3 / 2) tanθ = 2 + 2√3 tanθTherefore, Distance squared becomes:4 / cos²θ + 1 - [2 + 2√3 tanθ] = 4 / cos²θ + 1 - 2 - 2√3 tanθ = 4 / cos²θ - 1 - 2√3 tanθHmm, that seems complicated. Let me write everything in terms of sinθ and cosθ.We know that 1 / cos²θ = 1 + tan²θ, so:4 / cos²θ = 4 (1 + tan²θ)So:Distance squared = 4(1 + tan²θ) - 1 - 2√3 tanθ = 4 + 4 tan²θ - 1 - 2√3 tanθ = 3 + 4 tan²θ - 2√3 tanθLet me denote t = tanθ. Then the expression becomes:3 + 4t² - 2√3 tWhich is a quadratic in t: 4t² - 2√3 t + 3To find the minimum of this quadratic, since the coefficient of t² is positive, the minimum occurs at t = -b/(2a) = (2√3)/(2*4) = √3/4So the minimum value is 4*(√3/4)^2 - 2√3*(√3/4) + 3Calculate:4*(3/16) - 2√3*(√3/4) + 3 = (12/16) - (6/4) + 3 = (3/4) - (3/2) + 3 = (3/4 - 3/2) + 3 = (-3/4) + 3 = 9/4Therefore, the minimum distance squared is 9/4, so the minimum distance is 3/2, which matches the previous result.So both methods give the same answer, 3/2.But let me check again. When I converted to Cartesian coordinates, the minimum distance was 3/2, which is 1.5. That seems correct because the horizontal distance from x=0.5 to x=2 is 1.5, and since the closest point is directly horizontal (same y-coordinate), the distance is purely horizontal. So that's why when we set y = √3/2, the vertical component difference is zero, so distance is 1.5.Therefore, both methods confirm that the minimum distance is 3/2.But let me just make sure there's no mistake in the polar coordinate method. Let's go through the steps again.We had to compute the distance squared between P(ρ, θ) on ρ = 2 secθ and Q(1, π/3). The formula in polar coordinates is sqrt(ρ² + 1 - 2ρ*1*cos(θ - π/3)). Then, substituting ρ = 2 secθ gives:Distance squared = 4 sec²θ + 1 - 4 secθ cos(θ - π/3)Then, expanding cos(θ - π/3) as 0.5 cosθ + (√3/2) sinθ, leading to:4 secθ [0.5 cosθ + (√3/2) sinθ] = 4 * 0.5 + 4 * (√3/2) tanθ = 2 + 2√3 tanθHence, Distance squared = 4 sec²θ +1 - 2 - 2√3 tanθ = 4 sec²θ -1 -2√3 tanθThen, using 1 + tan²θ = sec²θ, so 4 sec²θ = 4 + 4 tan²θ, so substituting:4 + 4 tan²θ -1 -2√3 tanθ = 3 + 4 tan²θ -2√3 tanθThen, substituting t = tanθ:3 +4t² -2√3 tThis is a quadratic in t, which has its minimum at t = (2√3)/(2*4) = √3/4Plugging back into the quadratic:4*(√3/4)^2 -2√3*(√3/4) +3Compute:4*(3/16) - (2√3*√3)/4 +3 = 12/16 - 6/4 +3 = 3/4 - 3/2 +3 = (-3/4) +3 = 9/4So indeed, the minimum distance squared is 9/4, so the distance is 3/2. Perfect, that's consistent.Alternatively, using calculus in polar coordinates might be another approach. Let me try that as well for thoroughness.Expressing the distance squared as a function of θ:D² = ρ² + 1 - 2ρ cos(θ - π/3)But since ρ = 2 secθ, substitute:D² = (4 sec²θ) + 1 - 2*(2 secθ)*cos(θ - π/3)Which is the same as before. Then taking derivative with respect to θ and setting to zero. But that might be more complicated. However, since we already solved it by substitution and got the answer, maybe it's redundant. But just to check, let's compute the derivative.Let D² = 4 sec²θ + 1 - 4 secθ cos(θ - π/3)Let’s compute d(D²)/dθ:First term: derivative of 4 sec²θ is 4 * 2 sec²θ tanθ = 8 sec²θ tanθSecond term: derivative of 1 is 0Third term: derivative of -4 secθ cos(θ - π/3). Let's use product rule.Let’s denote u = secθ and v = cos(θ - π/3). Then the derivative is -4 [u’ v + u v’]Compute u’ = derivative of secθ = secθ tanθv’ = derivative of cos(θ - π/3) = -sin(θ - π/3)Therefore:-4 [ secθ tanθ * cos(θ - π/3) + secθ * (-sin(θ - π/3)) ]= -4 [ secθ tanθ cos(θ - π/3) - secθ sin(θ - π/3) ]Factor out secθ:= -4 secθ [ tanθ cos(θ - π/3) - sin(θ - π/3) ]Now, set the entire derivative equal to zero:8 sec²θ tanθ -4 secθ [ tanθ cos(θ - π/3) - sin(θ - π/3) ] = 0Divide both sides by 4 secθ (assuming secθ ≠ 0, which is true since ρ = 2 secθ would require cosθ ≠ 0):2 secθ tanθ - [ tanθ cos(θ - π/3) - sin(θ - π/3) ] = 0Let me write secθ as 1/cosθ:2 (1/cosθ) tanθ - [ tanθ cos(θ - π/3) - sin(θ - π/3) ] = 0Note that tanθ = sinθ / cosθ, so:2 (sinθ / cos²θ) - [ (sinθ / cosθ) cos(θ - π/3) - sin(θ - π/3) ] = 0Simplify the second term:(sinθ / cosθ) cos(θ - π/3) = sinθ cos(θ - π/3) / cosθSo:2 (sinθ / cos²θ) - [ sinθ cos(θ - π/3)/cosθ - sin(θ - π/3) ] = 0Let me factor out sinθ / cosθ from the first term inside the brackets:= 2 (sinθ / cos²θ) - [ (sinθ / cosθ)(cos(θ - π/3)) - sin(θ - π/3) ] = 0This seems complicated. Maybe there's a trigonometric identity to simplify this expression.Alternatively, let me consider substituting φ = θ - π/3, which might make the expression easier. Let me set φ = θ - π/3, so θ = φ + π/3. Then, perhaps we can express everything in terms of φ.But this might not necessarily simplify things. Alternatively, let me consider expanding cos(θ - π/3) and sin(θ - π/3) using angle subtraction formulas.Recall that:cos(θ - π/3) = cosθ cosπ/3 + sinθ sinπ/3 = 0.5 cosθ + (√3/2) sinθsin(θ - π/3) = sinθ cosπ/3 - cosθ sinπ/3 = 0.5 sinθ - (√3/2) cosθSo substituting these into the expression:First, let's rewrite the bracketed term:sinθ cos(θ - π/3)/cosθ - sin(θ - π/3)= [sinθ (0.5 cosθ + (√3/2) sinθ)/cosθ] - [0.5 sinθ - (√3/2) cosθ]= [0.5 sinθ + (√3/2) sin²θ / cosθ] - 0.5 sinθ + (√3/2) cosθ= (0.5 sinθ - 0.5 sinθ) + (√3/2)(sin²θ / cosθ + cosθ)= 0 + (√3/2)( (sin²θ + cos²θ)/cosθ )Since sin²θ + cos²θ = 1:= (√3/2)(1 / cosθ )Therefore, the entire expression becomes:2 (sinθ / cos²θ) - [ (√3/2)(1 / cosθ ) ] = 0Multiply through by cos²θ to eliminate denominators:2 sinθ - (√3/2) cosθ = 0Multiply both sides by 2:4 sinθ - √3 cosθ = 0So:4 sinθ = √3 cosθDivide both sides by cosθ:4 tanθ = √3Therefore:tanθ = √3 /4So θ = arctan(√3 /4 )But this is the critical point. Then, let me compute tanθ = √3 /4. So θ is an angle whose tangent is √3 /4. Let me compute the value of θ. But maybe we don't need to compute θ numerically. Instead, let's compute tanθ = √3 /4. So in the quadratic approach earlier, we set t = tanθ and found the minimum occurs at t = √3 /4, which matches this result. Therefore, θ = arctan(√3 /4 ). Then, substituting back into the distance squared expression.But we already did that substitution and found the minimum distance squared is 9/4, so distance is 3/2. Therefore, this calculus approach also leads to the same conclusion.Therefore, regardless of the method—Cartesian conversion, substitution in polar coordinates, calculus in polar coordinates—the minimum distance is 3/2.But just to ensure completeness, let's verify the point of closest approach. In Cartesian, the closest point on x=2 to (0.5, √3/2) is (2, √3/2). Let's convert this back to polar coordinates to see if it lies on the original curve.Convert (2, √3/2) to polar coordinates. The radius ρ is sqrt(2² + (√3/2)^2) = sqrt(4 + 3/4) = sqrt(19/4) = √19 / 2 ≈ 2.179. The angle θ is arctan( (√3/2)/2 ) = arctan(√3/4 ), which matches the earlier result θ = arctan(√3/4). So, in polar coordinates, this point is (√19 / 2, arctan(√3/4 )). Let's check if this point lies on the curve ρ² cosθ - 2ρ =0.Compute ρ² cosθ - 2ρ:( (√19 / 2 )² ) cosθ - 2*(√19 / 2 )= (19/4) cosθ - √19But cosθ = 2 / ρ, since from the curve equation ρ = 2 secθ => cosθ = 2 / ρ. Therefore:(19/4)*(2 / (√19 / 2)) - √19= (19/4)*(4 / √19 ) - √19= (19/4)*(4 / √19 ) - √19= (19 / √19 ) - √19= √19 - √19 = 0Hence, the point indeed lies on the curve. Therefore, everything checks out.Thus, after multiple approaches and thorough verification, the minimum distance from point P on the curve ρ² cosθ - 2ρ =0 to point Q(1, π/3) is 3/2.Final AnswerThe minimum distance is boxed{dfrac{3}{2}}.</think>Given that point ( P ) moves on the curve ( rho^2 cos theta - 2rho = 0 ) in the polar coordinate system, we need to find the minimum distance from point ( P ) to point ( Q(1, frac{pi}{3}) ).First, we rewrite the given curve equation:[rho^2 cos theta - 2rho = 0 implies rho(rho cos theta - 2) = 0]Since ( rho neq 0 ), the curve simplifies to ( rho = 2 sec theta ), which is the vertical line ( x = 2 ) in Cartesian coordinates.Point ( Q(1, frac{pi}{3}) ) in Cartesian coordinates is:[x = 1 cos frac{pi}{3} = 0.5, quad y = 1 sin frac{pi}{3} = frac{sqrt{3}}{2}]Thus, ( Q ) is at ( (0.5, frac{sqrt{3}}{2}) ).The minimum distance from the vertical line ( x = 2 ) to the point ( (0.5, frac{sqrt{3}}{2}) ) is the horizontal distance between ( x = 2 ) and ( x = 0.5 ), which is ( |2 - 0.5| = 1.5 ).To confirm, we use the distance formula in polar coordinates. The distance squared between ( P(rho, theta) ) on ( rho = 2 sec theta ) and ( Q(1, frac{pi}{3}) ) is:[rho^2 + 1 - 2rho cos(theta - frac{pi}{3})]Substituting ( rho = 2 sec theta ):[4 sec^2 theta + 1 - 4 sec theta cos(theta - frac{pi}{3})]Using trigonometric identities and simplifying, we find the minimum value of the quadratic expression in ( tan theta ) leads to the minimum distance squared as ( frac{9}{4} ), hence the minimum distance is ( frac{3}{2} ).Thus, the minimum distance from point ( P ) to point ( Q ) is boxed{dfrac{3}{2}}.]

🔑:First, convert the polar coordinates of point Q(1, frac{pi}{3}) to rectangular coordinates, which gives us Q(frac{1}{2}, frac{sqrt{3}}{2}).Next, transform the polar equation of the curve, rho^{2}cos theta-2rho=0, into rectangular coordinates. This results in the equation x=2.Since the curve is a vertical line at x=2, the minimum distance from any point P on this curve to point Q(frac{1}{2}, frac{sqrt{3}}{2}) would be the absolute difference between their x-coordinates.Therefore, the minimum distance from point P to point Q is 2 - frac{1}{2} = boxed{frac{3}{2}}.

🧠:[<think>Okay, let me try to figure out how to solve this problem. So, the expression is:20130 + 2*(480*4*14 + 30*44*16) - (5*80*43 + 19*400*3)*2Hmm, that looks a bit complicated with all the multiplications and additions. Let me break it down step by step. Maybe I can tackle each part separately and then combine them at the end. First, let me look at the structure of the expression. It's 20130 plus 2 times some terms in parentheses, then minus another parentheses term multiplied by 2. So, order of operations is important here. I need to compute the products inside each parentheses first, then multiply by the coefficients outside, and finally add and subtract everything together.Starting with the first part: 2*(480*4*14 + 30*44*16). Let me compute each multiplication inside the parentheses separately.First term inside the first parentheses: 480*4*14. Let me compute 480*4 first. 480 times 4. 400*4 is 1600, and 80*4 is 320, so 1600+320=1920. Then, multiply that by 14. 1920*14. Hmm, how to do that. Let's break 14 into 10 + 4. So, 1920*10=19200, 1920*4=7680. Adding those together: 19200+7680=26880. So, 480*4*14=26,880.Now the second term inside the first parentheses: 30*44*16. Let me compute 30*44 first. 30*40=1200, 30*4=120, so 1200+120=1320. Then, 1320*16. Let me break that down. 1320*10=13,200, 1320*6=7,920. Adding those gives 13,200+7,920=21,120. So, 30*44*16=21,120.Adding those two results inside the parentheses: 26,880 + 21,120. Let's see, 26,880 + 21,120. 26,000 + 21,000=47,000. 880 + 120=1,000. So total is 47,000 +1,000=48,000. So the first part inside the parentheses is 48,000. Then multiply by 2: 48,000*2=96,000. So the first part after 20130 is +96,000.Now, moving on to the second part: -(5*80*43 +19*400*3)*2. Again, let me compute the terms inside the parentheses first.First term inside the second parentheses: 5*80*43. Let's compute 5*80 first, which is 400. Then, 400*43. Hmm, 400*40=16,000 and 400*3=1,200, so 16,000+1,200=17,200. So, 5*80*43=17,200.Second term inside the second parentheses: 19*400*3. Let's compute 19*400 first. 19*400 is 19*4*100=76*100=7,600. Then, multiply by 3: 7,600*3=22,800. So, 19*400*3=22,800.Adding those two results inside the parentheses: 17,200 +22,800. Let's see, 17,000 +22,000=39,000. 200 +800=1,000. So total is 39,000 +1,000=40,000. Therefore, the inside of the second parentheses is 40,000. Then multiply by 2: 40,000*2=80,000. But there's a negative sign in front, so it becomes -80,000.Now, putting it all together:Original expression is 20130 +96,000 -80,000.Let me compute that step by step. First, 20,130 +96,000. Let's add those. 20,000 +96,000=116,000. Then add the remaining 130: 116,000 +130=116,130.Then subtract 80,000: 116,130 -80,000=36,130.Wait, is that right? Let me verify the calculations again to be sure.First part: 480*4*14.480*4=1,920. Then 1,920*14. Let me compute 1,920*10=19,200; 1,920*4=7,680. 19,200+7,680=26,880. That's correct.30*44*16: 30*44=1,320; 1,320*16=21,120. Correct. Then 26,880+21,120=48,000. Multiply by 2 gives 96,000. Correct.Second part: 5*80*43. 5*80=400; 400*43=17,200. Correct. 19*400*3: 19*400=7,600; 7,600*3=22,800. Correct. Then 17,200+22,800=40,000. Multiply by 2 gives 80,000. But since there's a minus sign, it's -80,000.So, 20,130 +96,000=116,130; 116,130 -80,000=36,130. Hmm, seems correct.Wait, let me check the original problem again to make sure I copied it right. The original expression is:20130 +2*(480*4*14 +30*44*16) - (5*80*43 +19*400*3)*2Yes, that's correct. So the result is 36,130. Wait, but let me check once more.Wait, the first part is 20130 + [2*(...)] which is 96,000, so 20130 +96,000=116,130. Then subtract [ (sum)*2 ] which is 80,000, so 116,130 -80,000=36,130. That seems right.But let me confirm each step again, because sometimes even if you think you did it right, there might be a miscalculation.First parentheses:480*4*14=480*(4*14)=480*56. Wait, 4*14=56. So 480*56. Let me compute that. 480*50=24,000; 480*6=2,880. 24,000 +2,880=26,880. That's the same as before.30*44*16: 30*44=1,320. 1,320*16: 1,000*16=16,000; 320*16=5,120. 16,000+5,120=21,120. Correct.Adding those gives 48,000. Multiply by 2: 96,000. Correct.Second parentheses:5*80*43. 5*80=400. 400*43. Let me check 400*40=16,000; 400*3=1,200. Total 17,200. Correct.19*400*3. 19*400=7,600. 7,600*3=22,800. Correct. Sum is 40,000. Multiply by 2: 80,000. Correct.So 20130 +96,000=116,130 -80,000=36,130. So the answer is 36,130. Therefore, the final result is 36,130. But wait, let me check once more.Wait, maybe I made an error in adding 20130 and 96,000. Let's do that again:20,130 +96,000. 20,000 +96,000=116,000. Then add the 130: 116,000 +130=116,130. Then subtract 80,000: 116,130 -80,000. 116,130 -80,000=36,130. Yep, that's correct.Alternatively, maybe compute in another way. Let's write all the numbers:20,130 + (2*(480*4*14 +30*44*16)) - (2*(5*80*43 +19*400*3))Compute the first big term: 2*(480*4*14 +30*44*16) = 2*(26,880 +21,120)=2*(48,000)=96,000.Second big term: 2*(5*80*43 +19*400*3)=2*(17,200 +22,800)=2*(40,000)=80,000.So overall: 20,130 +96,000 -80,000=20,130 + (96,000 -80,000)=20,130 +16,000=36,130. Oh! Wait, maybe that's a simpler way. 96,000 -80,000=16,000. Then 20,130 +16,000=36,130. Yeah, same result. So that's correct.Therefore, the final answer is 36,130. Wait, but I need to make sure that in the original problem, the parentheses are correctly placed. Let me check again:Original expression: 20130 +2*(... + ...) - (... + ...)*2. So the structure is correct. The order of operations is followed: parentheses first, then multiplication, then addition and subtraction. So yes, the calculation is correct.Therefore, the result is 36,130. So I think that's the answer. But to be absolutely sure, let me recompute each multiplication once again.First parentheses: 480*4*14.480*4: 4*400=1600, 4*80=320, total 1920.1920*14: 1920*10=19200, 1920*4=7680, total 19200+7680=26880. Correct.30*44*16:30*44: 30*40=1200, 30*4=120, total 1320.1320*16: 1320*10=13200, 1320*6=7920, total 13200+7920=21120. Correct.Sum: 26880+21120=48000. Multiply by 2: 96000. Correct.Second parentheses:5*80*43:5*80=400. 400*43=17,200. Correct.19*400*3:19*400=7600. 7600*3=22,800. Correct.Sum: 17,200+22,800=40,000. Multiply by 2: 80,000. Correct.So, 20130 +96000=116130. 116130 -80000=36130.Yes, all steps check out. Therefore, the answer is 36,130. So, I think that's the final answer. Maybe I can try another approach, like using distributive properties, but given the numbers, breaking them down step by step is safer. Alternatively, maybe using calculator-like steps, but since I don't have a calculator, manual computation is necessary. But all steps have been verified multiple times, so I think it's safe to conclude the answer is 36,130.Final Answerboxed{36130}</think>To solve the given expression:20130 + 2 cdot (480 cdot 4 cdot 14 + 30 cdot 44 cdot 16) - (5 cdot 80 cdot 43 + 19 cdot 400 cdot 3) cdot 2we break it down step by step.1. Calculate the terms inside the first parentheses: - (480 cdot 4 cdot 14): [ 480 cdot 4 = 1920 1920 cdot 14 = 26880 ] - (30 cdot 44 cdot 16): [ 30 cdot 44 = 1320 1320 cdot 16 = 21120 ] - Sum the results: [ 26880 + 21120 = 48000 ] - Multiply by 2: [ 2 cdot 48000 = 96000 ]2. Calculate the terms inside the second parentheses: - (5 cdot 80 cdot 43): [ 5 cdot 80 = 400 400 cdot 43 = 17200 ] - (19 cdot 400 cdot 3): [ 19 cdot 400 = 7600 7600 cdot 3 = 22800 ] - Sum the results: [ 17200 + 22800 = 40000 ] - Multiply by 2: [ 2 cdot 40000 = 80000 ]3. Combine all parts of the expression: - Start with 20130, add 96000, then subtract 80000: [ 20130 + 96000 = 116130 116130 - 80000 = 36130 ]Thus, the final answer is:[boxed{36130}],<think>Okay, so I need to calculate this big expression: 20130 plus 2 times (480 times 4 times 14 plus 30 times 44 times 16) minus (5 times 80 times 43 plus 19 times 400 times 3) times 2. Hmm, that's a bit complex, but let me break it down step by step.First, I remember that when dealing with expressions like this, I should follow the order of operations—PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). So, let's tackle the parts inside the parentheses first. There are two main parentheses here: one being multiplied by 2 and the other being multiplied by 2 after the minus sign. Let's handle each part separately.Starting with the first part: 20130 + 2 * (480 * 4 * 14 + 30 * 44 * 16). The main parentheses here have two terms: 480*4*14 and 30*44*16. Let me compute each of these.First term: 480 * 4 * 14. Let's compute 480*4 first. 480*4 is... let's see, 400*4=1600 and 80*4=320, so 1600+320=1920. Then multiply that by 14. Hmm, 1920*14. Let me break that down. 1920*10=19200, 1920*4=7680. Adding those together: 19200 + 7680 = 26880. So the first term inside the first parentheses is 26,880.Second term: 30 * 44 *16. Let's compute 30*44 first. 30*40=1200 and 30*4=120, so 1200+120=1320. Then multiply that by 16. 1320*16. Let's do 1000*16=16,000, 300*16=4,800, 20*16=320. So 16,000 + 4,800 = 20,800 + 320 = 21,120. So the second term is 21,120.Now add those two results together: 26,880 + 21,120. Let's see, 26,880 + 21,120. 26,000 + 21,000 = 47,000. 880 + 120 = 1,000. So total is 47,000 + 1,000 = 48,000. Wait, that's convenient. So inside the first parentheses, the total is 48,000.Then multiply that by 2: 2 * 48,000 = 96,000. So the first part of the expression is 20,130 + 96,000. Let's compute that: 20,130 + 96,000. 20,000 + 96,000 = 116,000, plus 130 is 116,130. So that's the first part: 116,130.Now, moving on to the second part: -(5 * 80 * 43 + 19 * 400 * 3) * 2. Let's handle the parentheses here as well. Again, two terms inside: 5*80*43 and 19*400*3.First term: 5*80*43. Let's compute 5*80 first. 5*80=400. Then multiply by 43. 400*43. Let's calculate that: 400*40=16,000 and 400*3=1,200, so total is 16,000 + 1,200 = 17,200. So first term is 17,200.Second term: 19*400*3. Let's compute 19*400 first. 19*400 is 19*4*100=76*100=7,600. Then multiply by 3: 7,600*3=22,800. So second term is 22,800.Adding those two together: 17,200 + 22,800. Let's see, 17,000 + 22,000 = 39,000. 200 + 800 = 1,000. So total is 39,000 + 1,000 = 40,000. So inside the second parentheses, the total is 40,000.Then multiply that by 2: 40,000*2=80,000. But there's a negative sign in front, so it's -80,000.Now, putting it all together: the entire expression is the first part minus the second part. So 116,130 - 80,000. Let's compute that: 116,130 - 80,000. That's 36,130.Wait, let me double-check all my steps because that seems straightforward, but maybe I made a miscalculation somewhere.Starting with the first parentheses: 480*4*14. 480*4=1,920. 1,920*14. 1,920*10=19,200; 1,920*4=7,680. 19,200+7,680=26,880. Correct. Then 30*44*16. 30*44=1,320. 1,320*16. 1,000*16=16,000; 300*16=4,800; 20*16=320. 16,000+4,800=20,800; 20,800+320=21,120. Correct. Then 26,880 +21,120=48,000. Multiply by 2: 96,000. Then 20,130 +96,000=116,130. That's correct.Second parentheses: 5*80*43=5*80=400. 400*43=17,200. Then 19*400*3. 19*400=7,600. 7,600*3=22,800. Then 17,200 +22,800=40,000. Multiply by 2: 80,000. Then subtract that from the first part: 116,130 -80,000=36,130.Hmm, so the final answer is 36,130. Let me check once again to be sure. Maybe I should verify the multiplications again.First term inside first parentheses: 480*4=1,920. 1,920*14. Let me compute 1,920*10=19,200. 1,920*4=7,680. 19,200 +7,680=26,880. Correct.Second term:30*44=1,320. 1,320*16. Let's do 1,320*10=13,200. 1,320*6=7,920. 13,200 +7,920=21,120. Correct. 26,880 +21,120=48,000. Multiply by 2: 96,000. 20,130 +96,000=116,130. Correct.Second part:5*80=400. 400*43. 400*40=16,000. 400*3=1,200. Total 17,200. Correct. 19*400=7,600. 7,600*3=22,800. Correct. 17,200 +22,800=40,000. Multiply by 2:80,000. So subtract 80,000 from 116,130: 116,130 -80,000=36,130.Yes, seems right. Maybe I can check if there's another way to compute some parts to confirm.Alternatively, maybe compute the entire expression step by step again.Original expression:20130 + 2*(480*4*14 + 30*44*16) - (5*80*43 + 19*400*3)*2Compute each multiplication inside the first parentheses:480*4=1,920; 1,920*14=26,880.30*44=1,320; 1,320*16=21,120.Sum:26,880 +21,120=48,000.Multiply by 2:48,000*2=96,000.Add to 20130:20,130 +96,000=116,130.Now the second parentheses:5*80=400; 400*43=17,200.19*400=7,600; 7,600*3=22,800.Sum:17,200 +22,800=40,000.Multiply by 2:40,000*2=80,000.Subtract from previous total:116,130 -80,000=36,130.Same result. So I think that's correct.Alternatively, maybe I can check if 480*4*14 is indeed 26,880. 480*14=6,720. 6,720*4=26,880. Wait, 480*14: 400*14=5,600; 80*14=1,120. 5,600 +1,120=6,720. Then times 4 is 26,880. Correct.Similarly, 30*44*16:30*44=1,320. 1,320*16: 1,320*10=13,200; 1,320*6=7,920. Total 21,120. Correct.5*80*43:5*80=400. 400*43=17,200. 19*400*3:19*400=7,600. 7,600*3=22,800. Correct.So everything checks out. Therefore, the answer is 36,130.Final Answerboxed{36130}</think>To calculate the given expression:20130 + 2 cdot (480 cdot 4 cdot 14 + 30 cdot 44 cdot 16) - (5 cdot 80 cdot 43 + 19 cdot 400 cdot 3) cdot 2we break it down step by step.First, compute the terms inside the parentheses:1. Calculate (480 cdot 4 cdot 14): [ 480 cdot 4 = 1920 ] [ 1920 cdot 14 = 26880 ]2. Calculate (30 cdot 44 cdot 16): [ 30 cdot 44 = 1320 ] [ 1320 cdot 16 = 21120 ]3. Add the results: [ 26880 + 21120 = 48000 ]4. Multiply by 2: [ 2 cdot 48000 = 96000 ]5. Add this to 20130: [ 20130 + 96000 = 116130 ]Next, compute the terms inside the second parentheses:1. Calculate (5 cdot 80 cdot 43): [ 5 cdot 80 = 400 ] [ 400 cdot 43 = 17200 ]2. Calculate (19 cdot 400 cdot 3): [ 19 cdot 400 = 7600 ] [ 7600 cdot 3 = 22800 ]3. Add the results: [ 17200 + 22800 = 40000 ]4. Multiply by 2: [ 2 cdot 40000 = 80000 ]Finally, subtract this result from the previous total:[116130 - 80000 = 36130]Thus, the final answer is:[boxed{36130}]]

🔑:Solution. By direct calculation, we getbegin{aligned}20130+2 cdot(480 cdot 4 cdot 14 & +30 cdot 44 cdot 16)-(5 cdot 80 cdot 43+19 cdot 400 cdot 3) cdot 2= & =20130+2 cdot(480 cdot 56+44 cdot 480)-(400 cdot 43+400 cdot 57) cdot 2 & =20130+2 cdot 480 cdot 100-2 cdot 400 cdot 100 & =20130+96000-80000 & =20130+16000=36130end{aligned}

🧠:[<think>Okay, let me try to solve these equations one by one. Starting with equation (1). Hmm, the equation is (a + b)(a x + b)(a - b x) equals (a² x - b²)(a + b x). I need to solve for x, right? Let me expand both sides and see if I can simplify it.First, let's take the left-hand side (LHS): (a + b)(a x + b)(a - b x). Maybe I can multiply them step by step. Let me first multiply (a + b) and (a x + b). Let's distribute:(a + b)(a x + b) = a*(a x + b) + b*(a x + b) = a² x + a b + a b x + b². Combine like terms: a² x + a b x + a b + b². So that's a x(a + b) + b(a + b). Wait, that's (a x + b)(a + b). Hmm, maybe there's a better way to multiply these.Alternatively, maybe it's easier to multiply the first two factors first. Let's see:First multiply (a + b)(a x + b). Let me use the distributive property again:= a*(a x) + a*b + b*(a x) + b*b= a² x + a b + a b x + b²Combine terms: a² x + a b x + a b + b². Then, multiply this by (a - b x). That seems complicated, but let's try.So (a² x + a b x + a b + b²)*(a - b x). Let's distribute each term:First term: a² x*(a - b x) = a³ x - a² b x²Second term: a b x*(a - b x) = a² b x - a b² x²Third term: a b*(a - b x) = a² b - a b² xFourth term: b²*(a - b x) = a b² - b³ xNow, let's collect all these terms:= a³ x - a² b x² + a² b x - a b² x² + a² b - a b² x + a b² - b³ xNow, combine like terms:Terms with x²: -a² b x² - a b² x² = -a b x²(a + b)Terms with x: a³ x + a² b x - a b² x - b³ x = x(a³ + a² b - a b² - b³) = x(a²(a + b) - b²(a + b)) = x(a + b)(a² - b²) = x(a + b)^2(a - b)Constant terms (without x): a² b + a b² = a b(a + b)So overall, the LHS simplifies to:- a b x²(a + b) + x(a + b)^2(a - b) + a b(a + b)Factor out (a + b):= (a + b)[ -a b x² + x(a + b)(a - b) + a b ]Now let's look at the right-hand side (RHS): (a² x - b²)(a + b x)Multiply these two:= a² x*(a) + a² x*(b x) - b²*(a) - b²*(b x)= a³ x + a² b x² - a b² - b³ xSo RHS is a³ x + a² b x² - a b² - b³ xNow, set LHS equal to RHS:(a + b)[ -a b x² + x(a + b)(a - b) + a b ] = a³ x + a² b x² - a b² - b³ xBut let's also note that RHS can be written as a² b x² + a³ x - b³ x - a b². Let's see if we can factor RHS:a³ x - b³ x + a² b x² - a b² = x(a³ - b³) + a b(a x - b)But a³ - b³ = (a - b)(a² + a b + b²), so:= x(a - b)(a² + a b + b²) + a b(a x - b)Hmm, maybe not helpful. Let's instead expand the LHS and RHS and see if we can subtract RHS from both sides to set equation to zero.First, expand LHS:From earlier, LHS was:(a + b)[ -a b x² + x(a + b)(a - b) + a b ]Multiply out:= (a + b)*(-a b x²) + (a + b)*x(a + b)(a - b) + (a + b)*a b= -a b(a + b)x² + x(a + b)^2(a - b) + a b(a + b)So, the equation is:- a b(a + b)x² + x(a + b)^2(a - b) + a b(a + b) = a³ x + a² b x² - a b² - b³ xBring all terms to left side:- a b(a + b)x² + x(a + b)^2(a - b) + a b(a + b) - a³ x - a² b x² + a b² + b³ x = 0Combine like terms:First, x² terms:- a b(a + b)x² - a² b x² = -a b(a + b + a) x² = -a b(2a + b) x²? Wait, let's compute:Wait, -a b(a + b)x² - a² b x² = -a b(a + b)x² - a² b x² = -a b x² [ (a + b) + a ] = -a b x² (2a + b)Wait, no. Wait, -a b(a + b) - a² b = -a² b - a b² -a² b = -2a² b - a b²Wait, maybe I made a mistake here. Let me re-express:Original x² terms from LHS: -a b(a + b)x²From RHS when moving to left: -a² b x²So total x² terms: -a b(a + b)x² - a² b x² = -a b(a + b + a)x² = -a b(2a + b)x². Wait, is that correct?Wait, no. Wait, -a b(a + b) -a² b = -a² b -a b² -a² b = -2a² b -a b². So it's not factorable as such. Maybe we need to keep it as is.Similarly, x terms:From LHS: x(a + b)^2(a - b)From RHS moved to left: -a³ x + b³ xWait, original equation after moving all terms to left:... + x(a + b)^2(a - b) - a³ x - a² b x² + a b² + b³ x = 0So x terms: x(a + b)^2(a - b) - a³ x + b³ xLet me factor x:x[ (a + b)^2(a - b) - a³ + b³ ]Similarly, constants:From LHS: a b(a + b)From RHS moved to left: + a b²So constants: a b(a + b) + a b² = a b(a + b + b) = a b(a + 2b)So putting it all together:The equation is:[ -a b(a + b) -a² b ]x² + x[ (a + b)^2(a - b) - a³ + b³ ] + a b(a + 2b) = 0Simplify each bracket:First, x² coefficient:- a b(a + b) -a² b = -a² b -a b² -a² b = -2a² b -a b² = -a b(2a + b)x term coefficient:(a + b)^2(a - b) -a³ + b³First compute (a + b)^2(a - b):(a + b)^2 = a² + 2a b + b², so multiplied by (a - b):= (a² + 2a b + b²)(a - b) = a³ - a² b + 2a² b - 2a b² + a b² - b³ = a³ + a² b - a b² - b³Therefore, (a + b)^2(a - b) -a³ + b³ = [a³ + a² b - a b² - b³] -a³ + b³ = a² b - a b²So x coefficient is a² b - a b² = a b(a - b)Constant term: a b(a + 2b)Therefore, equation simplifies to:- a b(2a + b)x² + a b(a - b)x + a b(a + 2b) = 0Factor out a b:a b [ - (2a + b)x² + (a - b)x + (a + 2b) ] = 0So either a b = 0 or the quadratic in the bracket equals zero.Case 1: a b = 0If a b = 0, then either a = 0 or b = 0.If a = 0: Substitute into original equation.Original equation: (0 + b)(0*x + b)(0 - b x) = (0 - b²)(0 + b x)Left side: (b)(b)(-b x) = b²*(-b x) = -b³ xRight side: (-b²)(b x) = -b³ xTherefore, both sides are equal. So when a = 0, the equation holds for all x. But the problem says "solve the equation", so if a = 0, then x can be any real number.Similarly, if b = 0: Check original equation.Original equation: (a + 0)(a x + 0)(a - 0) = (a² x - 0)(a + 0 x)Left side: a*(a x)*a = a³ xRight side: a² x * a = a³ xAgain, both sides equal. So if b = 0, equation holds for all x. So in the case a b = 0, the equation is identity, so x can be any real number.But maybe the problem assumes that a and b are such that a b ≠ 0? If not specified, we need to consider all possibilities.But moving on, if a b ≠ 0, then we can divide both sides by a b:- (2a + b)x² + (a - b)x + (a + 2b) = 0Which is a quadratic equation in x:( - (2a + b) )x² + (a - b)x + (a + 2b) = 0Multiply both sides by -1 to make it more standard:(2a + b)x² - (a - b)x - (a + 2b) = 0Now, let's solve this quadratic equation for x. Let me write it as:(2a + b)x² - (a - b)x - (a + 2b) = 0Use quadratic formula:x = [ (a - b) ± sqrt( (a - b)^2 + 4*(2a + b)*(a + 2b) ) ] / [ 2*(2a + b) ]Compute discriminant D:D = (a - b)^2 + 4*(2a + b)*(a + 2b)First expand (a - b)^2 = a² - 2a b + b²Then compute 4*(2a + b)(a + 2b):First compute (2a + b)(a + 2b):= 2a*a + 2a*2b + b*a + b*2b = 2a² + 4a b + a b + 2b² = 2a² + 5a b + 2b²Multiply by 4: 8a² + 20a b + 8b²Therefore, D = a² - 2a b + b² + 8a² + 20a b + 8b² = (1 + 8)a² + (-2 + 20)a b + (1 + 8)b² = 9a² + 18a b + 9b² = 9(a² + 2a b + b²) = 9(a + b)^2Therefore, sqrt(D) = 3(a + b)So plug back into quadratic formula:x = [ (a - b) ± 3(a + b) ] / [ 2*(2a + b) ]Compute the two solutions:First solution with +:x = [ (a - b) + 3(a + b) ] / [2*(2a + b)] = [ a - b + 3a + 3b ] / [2*(2a + b)] = [4a + 2b]/[2*(2a + b)] = [2*(2a + b)]/[2*(2a + b)] = 1Second solution with -:x = [ (a - b) - 3(a + b) ] / [2*(2a + b)] = [ a - b - 3a - 3b ] / [2*(2a + b)] = [ -2a -4b ] / [2*(2a + b)] = [ -2(a + 2b) ] / [2*(2a + b) ] = [ - (a + 2b) ] / (2a + b )Therefore, solutions are x = 1 and x = - (a + 2b)/(2a + b)But we need to check if these solutions are valid in the original equation, especially if denominators are involved. However, in the original equation, denominators are not present, so these solutions are valid unless 2a + b = 0 in the second solution. Wait, but if 2a + b = 0, then the original quadratic equation would have a zero coefficient for x², but in the case when a b ≠ 0, we divided by a b and got the quadratic. If 2a + b = 0, then the quadratic becomes linear.Wait, let's check if 2a + b = 0. If 2a + b = 0, then in the quadratic equation (2a + b)x² - (a - b)x - (a + 2b) = 0 becomes 0x² - (a - b)x - (a + 2b) = 0, which is a linear equation: - (a - b)x - (a + 2b) = 0. Then solution is x = - (a + 2b)/( - (a - b)) = (a + 2b)/(a - b). But this case only occurs when 2a + b = 0. So maybe in the case when 2a + b ≠ 0, we have two solutions, otherwise one solution. But in our previous quadratic solution, we divided by 2a + b, so when 2a + b = 0, the quadratic formula is invalid. Therefore, we need to consider separately when 2a + b = 0.But since we assumed a b ≠ 0, if 2a + b = 0, then b = -2a. Then, substituting back into original equation, let's check if original equation holds for any x.Original equation: (a + b)(a x + b)(a - b x) = (a² x - b²)(a + b x)If b = -2a, then substitute b = -2a:Left side: (a - 2a)(a x - 2a)(a - (-2a)x) = (-a)(a x - 2a)(a + 2a x) = (-a)(a(x - 2))(a(1 + 2x)) = (-a)*a*a*(x - 2)(1 + 2x) = -a³ (x - 2)(1 + 2x)Right side: (a² x - (-2a)^2)(a + (-2a)x) = (a² x - 4a²)(a - 2a x) = a²(x - 4) * a(1 - 2x) = a³ (x - 4)(1 - 2x)Therefore, equation becomes:- a³ (x - 2)(1 + 2x) = a³ (x - 4)(1 - 2x)Assuming a ≠ 0 (since a b ≠ 0 implies a ≠ 0), divide both sides by a³:- (x - 2)(1 + 2x) = (x - 4)(1 - 2x)Multiply out both sides:Left side: - [x*1 + x*2x - 2*1 - 2*2x] = - [x + 2x² - 2 - 4x] = - [2x² -3x -2] = -2x² + 3x + 2Right side: [x*1 - x*2x -4*1 +4*2x] = [x -2x² -4 +8x] = -2x² +9x -4Set left = right:-2x² + 3x + 2 = -2x² +9x -4Subtract -2x² from both sides:3x +2 =9x -46x =6 => x=1Therefore, when 2a + b =0, the equation reduces to x=1. So in this case, even though the quadratic would have a division by zero, the original equation still gives x=1 as a solution.Therefore, overall, considering all cases:1. If a = 0 or b =0, then any x is a solution.2. If a ≠ 0 and b ≠0: - If 2a + b ≠0: solutions are x=1 and x= - (a + 2b)/(2a + b) - If 2a + b =0: solution is x=1But x=1 is always a solution even when 2a + b =0. So perhaps x=1 is always a solution regardless, and the other solution depends on the relationship between a and b.But let's verify with x=1. Let's check original equation with x=1:Left side: (a + b)(a*1 + b)(a - b*1) = (a + b)(a + b)(a - b) = (a + b)^2(a - b)Right side: (a²*1 - b²)(a + b*1) = (a² - b²)(a + b) = (a - b)(a + b)(a + b) = (a - b)(a + b)^2Therefore, both sides equal. So x=1 is always a solution, regardless of a and b.The other solution is x = - (a + 2b)/(2a + b). Let's check this solution in the original equation.Let me substitute x = - (a + 2b)/(2a + b) into original equation.Left side: (a + b)(a x + b)(a - b x)Compute a x + b: a*(-(a + 2b)/(2a + b)) + b = [ -a(a + 2b) + b(2a + b) ] / (2a + b) = [ -a² -2a b +2a b +b² ] / (2a + b) = (-a² + b²)/(2a + b) = -(a² - b²)/(2a + b) = -(a - b)(a + b)/(2a + b)Compute a - b x: a - b*(-(a + 2b)/(2a + b)) = [ a(2a + b) + b(a + 2b) ] / (2a + b ) = [2a² +a b +a b +2b² ] / (2a + b ) = [2a² +2a b +2b² ] / (2a + b ) = 2(a² +a b +b² ) / (2a + b )Therefore, left side becomes:(a + b) * [ -(a - b)(a + b)/(2a + b) ] * [ 2(a² +a b +b² ) / (2a + b ) ]Multiply them together:= (a + b) * (- (a - b)(a + b)) * 2(a² +a b +b² ) / [ (2a + b)^2 ]= -2(a + b)^2(a - b)(a² +a b +b² ) / (2a + b)^2Now, compute right side:(a² x - b²)(a + b x)First compute a² x - b²: a²*(-(a + 2b)/(2a + b)) - b² = [ -a²(a + 2b) - b²(2a + b) ] / (2a + b )= [ -a³ -2a² b -2a b² -b³ ] / (2a + b ) = - (a³ +2a² b +2a b² +b³ ) / (2a + b )Factor numerator: Let's see, a³ +b³ = (a + b)(a² -a b +b²), so:a³ + 2a² b +2a b² +b³ = (a³ +b³) +2a² b +2a b² = (a + b)(a² -a b +b²) + 2a b(a + b ) = (a + b)(a² -a b +b² +2a b ) = (a + b)(a² +a b +b² )Therefore, a² x -b² = - (a + b)(a² +a b +b² ) / (2a + b )Then compute a + b x: a + b*(-(a +2b)/(2a + b )) = [ a(2a + b ) -b(a +2b ) ] / (2a + b ) = [2a² +a b -a b -2b² ] / (2a + b ) = [2a² -2b² ] / (2a + b ) = 2(a² -b² ) / (2a + b ) = 2(a -b)(a + b ) / (2a + b )Therefore, right side is:[ - (a + b)(a² +a b +b² ) / (2a + b ) ] * [ 2(a -b)(a + b ) / (2a + b ) ]Multiply them:= -2(a + b)^2(a -b)(a² +a b +b² ) / (2a + b )^2Which is equal to left side. Therefore, x = - (a +2b )/(2a +b ) is indeed a valid solution.Therefore, the solutions are x=1 and x= - (a +2b)/(2a +b ), provided that 2a +b ≠0. If 2a +b =0, then the second solution is invalid, but in that case, the equation reduces to x=1.But wait, earlier when we considered 2a +b=0, even with x= - (a +2b)/(2a +b ), if 2a +b=0, then substituting b=-2a into x gives x= - (a +2*(-2a))/(2a + (-2a))= - (a -4a)/0 = - (-3a)/0, which is undefined. So in that case, only x=1 is valid.Hence, summarizing equation (1):- If a =0 or b=0, any real x is a solution.- If a ≠0 and b≠0: - If 2a +b ≠0, solutions are x=1 and x= - (a +2b)/(2a +b ) - If 2a +b =0, the only solution is x=1.But the problem statement says "solve the equations" without specifying constraints on a and b. So probably assuming a and b are such that denominators are not zero. So unless they specify, we can present the solutions as x=1 and x= - (a +2b)/(2a +b ), noting that if 2a +b=0, then x=1 is the only solution, and if a or b=0, all x are solutions. But since the problem is in the style of a standard algebra problem, perhaps assuming a and b are constants such that 2a +b ≠0 and a b≠0, so two solutions x=1 and x= - (a +2b)/(2a +b ). So I think that's the answer they expect.Moving on to equation (2): x^(1/3) + (2x -3)^(1/3) = [12(x -1)]^(1/3)This is a cubic equation involving cube roots. Let me denote cube roots as cr for shorthand.So, cr(x) + cr(2x -3) = cr(12(x -1))Let me denote a = cr(x), b=cr(2x -3). Then the equation is a + b = cr(12(x -1)).But maybe we can cube both sides to eliminate the cube roots. However, cubing a + b = c gives (a + b)^3 = c^3, which is a^3 + b^3 + 3ab(a + b) = c^3.But in our case, a^3 = x, b^3 =2x -3, c^3 =12(x -1)Therefore, cubing both sides:x + (2x -3) + 3*a*b*(a + b) =12(x -1)Simplify left side:x +2x -3 +3ab*(a + b) =3x -3 +3ab*(a + b)Set equal to right side:3x -3 +3ab*(a + b) =12x -12Bring all terms to left side:3x -3 +3ab*(a + b) -12x +12=0Simplify:-9x +9 +3ab*(a + b)=0Divide both sides by 3:-3x +3 +ab*(a + b)=0 => ab(a + b) =3x -3But recall that a + b = cr(12(x -1))= [12(x -1)]^(1/3). Let's denote c = [12(x -1)]^(1/3), so a + b =c.Also, note that ab = cr(x)*cr(2x -3) = [x(2x -3)]^(1/3)But maybe we can express ab in terms of c.Alternatively, let's substitute a + b =c into the equation ab*c =3x -3.So, ab*c =3x -3.But we also have:a^3 =xb^3=2x -3c^3=12(x -1)Also, note that c^3=12(x -1). Let's try to express x in terms of c^3:x= (c^3)/12 +1So x= (c³ +12)/12But maybe not helpful. Alternatively, express 3x -3 as 3(x -1)=3*(c³)/12= c³/4. Wait:From c³=12(x -1), then x -1= c³/12, so 3(x -1)=c³/4. Therefore, 3x -3= c³/4Therefore, ab*c =c³/4 => ab= c²/4But ab= [x(2x -3)]^(1/3)So [x(2x -3)]^(1/3)=c²/4Cube both sides:x(2x -3)= (c²/4)^3= c^6 / 64But c^3=12(x -1), so c^6= [12(x -1)]²=144(x -1)^2Therefore, x(2x -3)=144(x -1)^2 /64= (144/64)(x -1)^2= (9/4)(x -1)^2So equation becomes:x(2x -3)=(9/4)(x -1)^2Multiply both sides by 4 to eliminate denominator:4x(2x -3)=9(x -1)^2Expand left side:8x² -12xRight side:9(x² -2x +1)=9x² -18x +9Bring all terms to left:8x² -12x -9x² +18x -9=0Simplify:- x² +6x -9=0Multiply both sides by -1:x² -6x +9=0This factors as (x -3)^2=0, so x=3Check if x=3 is a solution to original equation.Original equation: cr(3) + cr(2*3 -3)= cr(12*(3 -1))Compute each term:cr(3)=3^(1/3)cr(6 -3)=cr(3)=3^(1/3)Right side: cr(12*2)=cr(24)=24^(1/3)So left side:3^(1/3) +3^(1/3)=2*3^(1/3)Right side:24^(1/3)= (8*3)^(1/3)=2*3^(1/3)Therefore, both sides equal 2*3^(1/3). So x=3 is a valid solution.But wait, when we cube both sides, sometimes extraneous solutions can be introduced, but in this case, x=3 works. Is there any other solution? The quadratic gave a double root at x=3, so maybe that's the only solution.But let's check if there could be other solutions. The original equation involves cube roots, which are defined for all real numbers, so we don't have domain restrictions. Let's test another value, say x=1.Left side:cr(1) + cr(2*1 -3)=1 + cr(-1)=1 + (-1)=0Right side:cr(12*(1 -1))=cr(0)=0So 0=0. Wait, x=1 also satisfies the equation? But according to our solution, x=3 is the only solution. What's happening here.Wait, when we cubed both sides, perhaps introduced an extraneous solution? Let's check.Wait, if x=1:Original equation:1^(1/3) + (2*1 -3)^(1/3)= [12(1 -1)]^(1/3)Left side:1 + (-1)^(1/3)=1 + (-1)=0Right side:0^(1/3)=0So 0=0, which is valid. So x=1 is also a solution. But in our algebra above, we ended up with only x=3. So why did x=1 not show up?Ah, perhaps when we made substitutions, we lost some solutions. Let's go back.Original equation: cr(x) + cr(2x -3) = cr(12(x -1))Let me denote A = cr(x), B=cr(2x -3), C=cr(12(x -1)). The equation is A + B = C.When we cube both sides:(A + B)^3 = C^3 => A³ + B³ +3AB(A + B)=C³But since A³ =x, B³=2x -3, C³=12(x -1)So substituting:x + (2x -3) +3AB*C=12(x -1)Which leads to 3AB*C=9x -9, then AB*C=3x -3But then we substituted AB= [x(2x -3)]^(1/3) and C= [12(x -1)]^(1/3)But in reality, AB= A*B= cr(x)*cr(2x -3)= [x(2x -3)]^(1/3). However, cube roots can be negative, so even if x(2x -3) is negative, the product of cube roots is the cube root of the product. So that step is okay.Then we have:[x(2x -3)]^(1/3)* [12(x -1)]^(1/3) =3x -3Which simplifies to:[12x(2x -3)(x -1)]^(1/3)=3x -3Then cube both sides:12x(2x -3)(x -1)=(3x -3)^3=27(x -1)^3So:12x(2x -3)(x -1) -27(x -1)^3=0Factor out (x -1):(x -1)[12x(2x -3) -27(x -1)^2 ]=0So either x=1, or 12x(2x -3) -27(x -1)^2=0Compute the second factor:12x(2x -3)=24x² -36x27(x -1)^2=27(x² -2x +1)=27x² -54x +27Thus, 24x² -36x -27x² +54x -27=0Combine like terms:-3x² +18x -27=0Divide by -3:x² -6x +9=0 => (x -3)^2=0 => x=3Therefore, solutions are x=1 and x=3. However, when we did previous substitution, we arrived only at x=3 because we divided by (x -1) when factoring? Wait no, when we factored (x -1), the equation became (x -1)*[...]=0, so x=1 is a solution, and [...] gives x=3. But in our previous steps, when we had:12x(2x -3)(x -1) =27(x -1)^3We could have moved everything to one side:12x(2x -3)(x -1) -27(x -1)^3=0Factor out (x -1):(x -1)[12x(2x -3) -27(x -1)^2]=0Which gives x=1 or the other equation. But in the previous approach, we divided both sides by (x -1), assuming x≠1, leading us to x=3. But actually, when we set up the equation AB*C=3x -3, if x=1, then 3x -3=0, so AB*C=0. Since C=cr(12(x -1))=cr(0)=0, so AB*0=0, which is true regardless of AB. So x=1 is a valid solution.Therefore, by factoring, we see x=1 and x=3 are solutions. But when we earlier went through the process, we lost x=1 because when we expressed AB*C=3x -3, and then substituted AB and C in terms of x, and then proceeded, but in reality, x=1 makes both sides zero regardless.Therefore, the solutions are x=1 and x=3. However, we need to verify both.We already checked x=3 works. x=1 also works. So why did our previous steps not show x=1? Let's check where we might have lost it.After substituting and getting to the equation:[ x(2x -3) ]^(1/3) * [12(x -1) ]^(1/3) =3x -3But if x=1, left side is [1*(-1)]^(1/3)*[0]^(1/3)= (-1)^(1/3)*0= -1*0=0Right side:3*1 -3=0. So 0=0, which holds. But when we cubed both sides, we get:[ x(2x -3)*12(x -1) ] = (3x -3)^3At x=1, both sides become 0=0. So x=1 is valid. The reason it didn't appear in the quadratic solution is because when we divided by (x -1) during factoring, we didn't consider x=1. But in reality, x=1 is a solution.Therefore, the solutions are x=1 and x=3.But wait, let's check x=1 again in original equation:x^(1/3) + (2x -3)^(1/3) = (12(x -1))^(1/3)x=1:Left side:1^(1/3) + (-1)^(1/3)=1 + (-1)=0Right side:0^(1/3)=0So 0=0, correct.x=3:Left side:3^(1/3) +3^(1/3)=2*3^(1/3)Right side:24^(1/3)=2*3^(1/3)So correct.But are there any other solutions? The equation ended up as (x -1)(x -3)^2=0, so x=1 and x=3 (double root). But since we're dealing with real cube roots, which are odd functions, so negative values are allowed.Wait, what if x=1 and x=3 are the only real solutions? Let's see.Suppose we consider another x, say x=0.Left side:0 + (-3)^(1/3)= - cube root of 3Right side:12(-1)^(1/3)=12*(-1)= -12. But cube root of -12 is - cube root of12. So left side is -3^(1/3)≈-1.442, right side is -12^(1/3)≈-2.289. Not equal.x=2:Left side:cr(2) + cr(1)= approx1.26 +1=2.26Right side:cr(12(1))=cr(12)=approx2.289. Not equal.x=4:Left side:cr(4)+cr(5)= approx1.587 +1.710=3.297Right side:cr(12*3)=cr(36)≈3.301. Close, but not equal. Not exact.Therefore, only x=1 and x=3 are solutions.Thus, the solutions to equation (2) are x=1 and x=3.But wait, when we solved the equation by substitution and factoring, we got x=1 and x=3. However, initially, when we derived the quadratic, we thought only x=3 was the solution, but factoring showed x=1 is also a solution. Therefore, answer is x=1 and x=3.Hence, problem (2) solutions are x=1 and x=3.Final Answer1. Solutions are boxed{1} and boxed{-dfrac{a + 2b}{2a + b}}.2. Solutions are boxed{1} and boxed{3}.</think> Summary1. Equation (1): The given equation is ((a + b)(a x + b)(a - b x) = (a^2 x - b^2)(a + b x)). - Step-by-Step Solution: - Expand both sides and simplify. - Factor out common terms and solve the resulting quadratic equation. - Consider special cases where (a = 0) or (b = 0), leading to any (x) being a solution. - For (a neq 0) and (b neq 0), solve the quadratic equation using the quadratic formula. - Verify solutions and check for special cases where denominators might be zero. Solutions: - If (a = 0) or (b = 0), any real (x) is a solution. - For (a neq 0) and (b neq 0): - Solutions are (x = 1) and (x = -frac{a + 2b}{2a + b}). - If (2a + b = 0), the only solution is (x = 1). Final Answer: [ boxed{1} quad text{and} quad boxed{-dfrac{a + 2b}{2a + b}} ]2. Equation (2): The given equation is (x^{frac{1}{3}} + (2x - 3)^{frac{1}{3}} = (12(x - 1))^{frac{1}{3}}). - Step-by-Step Solution: - Cube both sides to eliminate cube roots. - Substitute and simplify the resulting equation. - Factor out common terms and solve the quadratic equation. - Verify solutions by substituting back into the original equation. Solutions: - The solutions are (x = 1) and (x = 3). Final Answer: [ boxed{1} quad text{and} quad boxed{3} ],<think>Okay, so I have these two equations to solve. Let's start with the first one. The equation is (a + b)(a x + b)(a - b x) equals (a² x - b²)(a + b x). Hmm, I need to solve for x here. Let me see. Maybe I can expand both sides and then simplify to find x. Let's try expanding the left side first.First, multiply (a + b) and (a x + b). Let's do that step by step. Multiply a by each term in (a x + b): a * a x = a² x, and a * b = a b. Then multiply b by each term in (a x + b): b * a x = a b x, and b * b = b². So combining these, we get a² x + a b + a b x + b². Then combine like terms: a² x + a b x = a x(a + b), and a b + b² = b(a + b). So overall, (a + b)(a x + b) = (a + b)(a x + b x + b)? Wait, maybe I should just keep it as a² x + a b + a b x + b² for now.Now, multiply this result by (a - b x). Let's denote the first product as C = a² x + a b + a b x + b². Then C * (a - b x). Let's expand term by term:First term: a² x * a = a³ xSecond term: a² x * (-b x) = -a² b x²Third term: a b * a = a² bFourth term: a b * (-b x) = -a b² xFifth term: a b x * a = a² b xSixth term: a b x * (-b x) = -a b² x²Seventh term: b² * a = a b²Eighth term: b² * (-b x) = -b³ xSo combining all these:a³ x - a² b x² + a² b - a b² x + a² b x - a b² x² + a b² - b³ xNow, let's combine like terms:Terms with x²: -a² b x² - a b² x² = -a b x²(a + b)Terms with x: a³ x - a b² x + a² b x - b³ xLet's factor x: x(a³ - a b² + a² b - b³)Then constant terms: a² b + a b²Wait, let me check each term again to make sure I didn't miss any.Wait, the first term is a³ x.Then -a² b x².Then +a² b.Then -a b² x.Then +a² b x.Then -a b² x².Then +a b².Then -b³ x.So combining the x² terms: -a² b x² - a b² x² = -a b x²(a + b)For x terms: a³ x - a b² x + a² b x - b³ x. Let's factor x:x(a³ + a² b - a b² - b³)Notice that a³ + a² b - a b² - b³ can be factored as a²(a + b) - b²(a + b) = (a² - b²)(a + b) = (a - b)(a + b)^2. Wait, is that right?Wait, let's compute a³ + a² b - a b² - b³:Group terms: (a³ + a² b) + (-a b² - b³) = a²(a + b) - b²(a + b) = (a² - b²)(a + b) = (a - b)(a + b)(a + b) = (a - b)(a + b)^2. Yeah, that's correct.So the x terms become x*(a - b)(a + b)^2.Constant terms: a² b + a b² = a b(a + b).So putting it all together, left side is:Left Side = -a b x²(a + b) + x*(a - b)(a + b)^2 + a b(a + b)Now, factor out (a + b):Left Side = (a + b)[ -a b x² + x*(a - b)(a + b) + a b ]Okay, now let's work on the right side: (a² x - b²)(a + b x)Multiply these two terms:First, a² x * a = a³ xThen a² x * b x = a² b x²Then -b² * a = -a b²Then -b² * b x = -b³ xSo combining terms:a³ x + a² b x² - a b² - b³ xSo the right side is:a³ x + a² b x² - a b² - b³ xLet me arrange terms similarly to the left side:Right Side = a² b x² + a³ x - b³ x - a b²Factor terms:Terms with x²: a² b x²Terms with x: a³ x - b³ x = x(a³ - b³) = x(a - b)(a² + a b + b²)Constant term: -a b²But maybe it's better to factor as follows:Right Side = a² b x² + x(a³ - b³) - a b²But let's see if we can factor further. Alternatively, let's see if we can write the right side in terms similar to the left side.But perhaps instead of expanding both sides, maybe we can subtract the right side from both sides and set the equation to zero, then factor.So let's compute Left Side - Right Side = 0.Left Side was:(a + b)[ -a b x² + x*(a - b)(a + b) + a b ]Right Side was:a² b x² + a³ x - b³ x - a b²Therefore, Left Side - Right Side:(a + b)[ -a b x² + x*(a - b)(a + b) + a b ] - (a² b x² + a³ x - b³ x - a b²) = 0Let me compute each part step by step.First, expand the left part:(a + b)*(-a b x²) + (a + b)*x*(a - b)(a + b) + (a + b)*a b= -a b (a + b) x² + x*(a - b)(a + b)^2 + a b(a + b)Now subtract the right side:- [a² b x² + a³ x - b³ x - a b²]So:- a² b x² - a³ x + b³ x + a b²Putting all together:Left Side - Right Side =[ -a b (a + b) x² + x*(a - b)(a + b)^2 + a b(a + b) ] - a² b x² - a³ x + b³ x + a b² = 0Now, combine like terms:First, terms with x²:- a b (a + b) x² - a² b x² = -a b(a + b + a) x² = -a b(2a + b) x² ? Wait, let me compute:Wait, -a b(a + b) x² - a² b x² = -a² b x² - a b² x² - a² b x² = (-a² b - a² b) x² - a b² x² = -2a² b x² - a b² x² = -a b x²(2a + b). Hmm, maybe not. Wait:Wait, actually, expanding:First term: -a b(a + b) x² = -a² b x² - a b² x²Second term: -a² b x²Therefore, combining them: (-a² b x² - a b² x²) + (-a² b x²) = -a² b x² - a² b x² - a b² x² = -2a² b x² - a b² x² = -a b x²(2a + b). Yeah, that's correct.Next, terms with x:x*(a - b)(a + b)^2 - a³ x + b³ xLet's factor x:x[ (a - b)(a + b)^2 - a³ + b³ ]Compute (a - b)(a + b)^2:First, (a + b)^2 = a² + 2a b + b²Multiply by (a - b):(a - b)(a² + 2a b + b²) = a(a² + 2a b + b²) - b(a² + 2a b + b²) = a³ + 2a² b + a b² - a² b - 2a b² - b³ = a³ + (2a² b - a² b) + (a b² - 2a b²) - b³ = a³ + a² b - a b² - b³So, (a - b)(a + b)^2 = a³ + a² b - a b² - b³Then, subtract a³ and add b³:So x[ (a³ + a² b - a b² - b³ ) - a³ + b³ ] = x[ a² b - a b² ]Therefore, x(a² b - a b²) = x a b(a - b)Then the constant terms:a b(a + b) + a b² = a b(a + b) + a b² = a² b + a b² + a b² = a² b + 2a b²So putting all together:Left Side - Right Side = -a b(2a + b) x² + x a b(a - b) + a² b + 2a b² = 0Factor out a b:a b [ - (2a + b) x² + (a - b) x + a + 2b ] = 0So, either a b = 0, or the quadratic in x is zero.Case 1: a b = 0.If a = 0, then substitute back into original equation:Left Side: (0 + b)(0*x + b)(0 - b x) = b * b * (-b x) = -b³ xRight Side: (0² x - b²)(0 + b x) = (-b²)(b x) = -b³ xSo equation becomes -b³ x = -b³ x, which is always true for any x, provided that b ≠ 0. But if a = 0 and b ≠ 0, then x can be any real number. However, if a = 0 and b = 0, then both sides are 0, so x is arbitrary. But in the original problem, since it's an equation to solve, perhaps a and b are parameters, not variables, so if a = 0 or b = 0, the equation might be identity or have infinitely many solutions. But unless specified, maybe we need to consider a and b ≠ 0. Maybe the problem assumes a and b are non-zero, so a b ≠ 0. So perhaps Case 1 is trivial, and we can proceed to Case 2.Case 2: - (2a + b) x² + (a - b) x + a + 2b = 0Multiply both sides by -1 to make it easier:(2a + b) x² - (a - b) x - (a + 2b) = 0Now, solve quadratic equation for x:(2a + b) x² - (a - b) x - (a + 2b) = 0Let me compute the discriminant D:D = [ - (a - b) ]² - 4*(2a + b)*(- (a + 2b)) = (a - b)^2 + 4(2a + b)(a + 2b)Expand (a - b)^2: a² - 2a b + b²Compute 4(2a + b)(a + 2b):First compute (2a + b)(a + 2b) = 2a*a + 2a*2b + b*a + b*2b = 2a² + 4a b + a b + 2b² = 2a² + 5a b + 2b²Multiply by 4: 8a² + 20a b + 8b²Then D = a² - 2a b + b² + 8a² + 20a b + 8b² = (a² + 8a²) + (-2a b + 20a b) + (b² + 8b²) = 9a² + 18a b + 9b² = 9(a² + 2a b + b²) = 9(a + b)^2So discriminant D = 9(a + b)^2Therefore, solutions are:x = [ (a - b) ± sqrt(D) ] / [2*(2a + b)] = [ (a - b) ± 3(a + b) ] / [2*(2a + b)]Compute both possibilities:First solution: [ (a - b) + 3(a + b) ] / [2*(2a + b) ] = [ a - b + 3a + 3b ] / [2*(2a + b) ] = [4a + 2b]/[2*(2a + b)] = 2*(2a + b)/[2*(2a + b)] = 1Second solution: [ (a - b) - 3(a + b) ] / [2*(2a + b) ] = [ a - b - 3a - 3b ] / [2*(2a + b) ] = [ -2a -4b ] / [2*(2a + b) ] = -2(a + 2b)/[2*(2a + b) ] = - (a + 2b)/(2a + b)Therefore, the solutions are x = 1 and x = - (a + 2b)/(2a + b)But we need to check if these solutions are valid in the original equation, in case of any restrictions.But since the original equation is a polynomial equation, and unless there's division by zero in the original equation (which there isn't, as the equation is polynomial), both solutions should be valid. However, let's verify.Check x = 1:Left Side: (a + b)(a*1 + b)(a - b*1) = (a + b)(a + b)(a - b) = (a + b)^2(a - b)Right Side: (a²*1 - b²)(a + b*1) = (a² - b²)(a + b) = (a - b)(a + b)(a + b) = (a - b)(a + b)^2Therefore, both sides are equal, so x = 1 is valid.Check x = - (a + 2b)/(2a + b):Let me compute Left Side and Right Side.First, denote x = - (a + 2b)/(2a + b). Let's compute each term step by step.Compute (a + b):(a + b)Compute (a x + b):a x + b = a*(- (a + 2b)/(2a + b)) + b = [ -a(a + 2b) + b(2a + b) ] / (2a + b) = [ -a² - 2a b + 2a b + b² ] / (2a + b) = (-a² + b²)/(2a + b) = -(a² - b²)/(2a + b) = -(a - b)(a + b)/(2a + b)Compute (a - b x):a - b x = a - b*(- (a + 2b)/(2a + b)) = a + b(a + 2b)/(2a + b) = [ a(2a + b) + b(a + 2b) ] / (2a + b) = [ 2a² + a b + a b + 2b² ] / (2a + b) = (2a² + 2a b + 2b²)/(2a + b) = 2(a² + a b + b²)/(2a + b)Multiply all three terms:Left Side = (a + b) * [ -(a - b)(a + b)/(2a + b) ] * [ 2(a² + a b + b²)/(2a + b) ]Multiply the constants:= (a + b) * (- (a - b)(a + b)) * 2(a² + a b + b²) / (2a + b)^2= -2(a + b)^2(a - b)(a² + a b + b²) / (2a + b)^2Now compute the Right Side:(a² x - b²)(a + b x)First compute a² x:a² x = a²*(- (a + 2b)/(2a + b)) = -a²(a + 2b)/(2a + b)Then subtract b²: a² x - b² = -a²(a + 2b)/(2a + b) - b² = [ -a²(a + 2b) - b²(2a + b) ] / (2a + b)= [ -a³ - 2a² b - 2a b² - b³ ] / (2a + b)Factor numerator:= - [ a³ + 2a² b + 2a b² + b³ ]Hmm, can this be factored? Let's see:a³ + 2a² b + 2a b² + b³ = a³ + b³ + 2a² b + 2a b² = (a + b)(a² - a b + b²) + 2a b(a + b) = (a + b)(a² - a b + b² + 2a b) = (a + b)(a² + a b + b²)So numerator is - (a + b)(a² + a b + b²)Therefore, a² x - b² = - (a + b)(a² + a b + b²)/(2a + b)Then compute (a + b x):a + b x = a + b*(- (a + 2b)/(2a + b)) = a - b(a + 2b)/(2a + b) = [ a(2a + b) - b(a + 2b) ] / (2a + b) = [ 2a² + a b - a b - 2b² ] / (2a + b) = (2a² - 2b²)/(2a + b) = 2(a² - b²)/(2a + b) = 2(a - b)(a + b)/(2a + b)Multiply both terms:Right Side = [ - (a + b)(a² + a b + b²)/(2a + b) ] * [ 2(a - b)(a + b)/(2a + b) ]= -2(a + b)^2(a - b)(a² + a b + b²)/(2a + b)^2Which is exactly equal to the Left Side. Therefore, x = - (a + 2b)/(2a + b) is also a valid solution.Therefore, the solutions to the first equation are x = 1 and x = - (a + 2b)/(2a + b).Now, moving on to the second equation: x^{1/3} + (2x - 3)^{1/3} = [12(x - 1)]^{1/3}This is a radical equation involving cube roots. Let me recall that cube roots are defined for all real numbers, so we don't have restrictions like with square roots, but we still need to check solutions in case of extraneous roots.Let me denote:Let’s set u = x^{1/3}, v = (2x - 3)^{1/3}, then the equation becomes u + v = [12(x - 1)]^{1/3}But perhaps it's better to cube both sides to eliminate the cube roots. Let's try that.First, let me write the equation as:∛x + ∛(2x - 3) = ∛[12(x - 1)]Let me denote left side as L = ∛x + ∛(2x - 3), right side as R = ∛[12(x - 1)]Cube both sides:L³ = R³So, (∛x + ∛(2x - 3))³ = [12(x - 1)]Expand the left side using the formula (a + b)^3 = a³ + b³ + 3ab(a + b):So,x + (2x - 3) + 3*∛x*∛(2x - 3)*(∛x + ∛(2x - 3)) = 12(x - 1)Simplify left side:x + 2x - 3 + 3*∛[x(2x - 3)]*(L) = 12x - 12Combine like terms:3x - 3 + 3*∛[2x² - 3x]*L = 12x - 12Subtract 3x - 3 from both sides:3*∛[2x² - 3x]*L = 12x - 12 - 3x + 3 = 9x - 9Divide both sides by 3:∛[2x² - 3x]*L = 3x - 3But L is the original left side, which is ∛x + ∛(2x - 3) = ∛[12(x - 1)]So, substitute back L = ∛[12(x - 1)]:Thus,∛[2x² - 3x] * ∛[12(x - 1)] = 3x - 3Note that ∛a * ∛b = ∛(ab), so:∛[ (2x² - 3x)*12(x - 1) ] = 3x - 3Compute the product inside the cube root:(2x² - 3x)*12(x - 1) = 12(x - 1)(2x² - 3x) = 12[2x³ - 3x² - 2x² + 3x] = 12[2x³ - 5x² + 3x] = 24x³ - 60x² + 36xTherefore, equation becomes:∛(24x³ - 60x² + 36x) = 3x - 3Cube both sides again to eliminate the cube root:24x³ - 60x² + 36x = (3x - 3)^3Compute the right side:(3x - 3)^3 = 27x³ - 81x² + 81x - 27So, equation:24x³ - 60x² + 36x = 27x³ - 81x² + 81x - 27Bring all terms to left side:24x³ - 60x² + 36x - 27x³ + 81x² - 81x + 27 = 0Combine like terms:(24x³ - 27x³) + (-60x² + 81x²) + (36x - 81x) + 27 = 0= (-3x³) + 21x² + (-45x) + 27 = 0Multiply both sides by -1 to make the leading coefficient positive:3x³ - 21x² + 45x - 27 = 0Divide both sides by 3:x³ - 7x² + 15x - 9 = 0Now, let's factor this cubic equation. Let's try rational roots. Possible rational roots are factors of 9 over factors of 1, so ±1, ±3, ±9.Test x = 1:1 - 7 + 15 - 9 = 0 → 0. So x = 1 is a root.Therefore, factor out (x - 1):Use polynomial division or synthetic division.Divide x³ -7x² +15x -9 by (x -1):Using synthetic division:1 | 1 -7 15 -9 1 -6 9 1 -6 9 0So, quotient polynomial is x² -6x +9, which factors as (x -3)^2.Therefore, x³ -7x² +15x -9 = (x -1)(x -3)^2Hence, roots are x =1 and x =3 (double root).Now, we need to check these solutions in the original equation to ensure they are valid, as we cubed both sides twice, which can introduce extraneous roots.Check x =1:Left side: ∛1 + ∛(2*1 -3) = 1 + ∛(-1) = 1 -1 = 0Right side: ∛[12(1 -1)] = ∛0 = 0So, 0 =0, valid.Check x =3:Left side: ∛3 + ∛(2*3 -3) = ∛3 + ∛3 = 2∛3Right side: ∛[12(3 -1)] = ∛[12*2] = ∛24 = 2∛3So, 2∛3 =2∛3, valid.Therefore, both x=1 and x=3 are solutions. Wait, but the cubic had x=3 as a double root. Let me check if x=3 is actually a valid solution, which it is, as above. So both roots are valid.Hence, solutions are x=1 and x=3.But let's check if there could be other solutions. Since cube roots are defined for all real numbers, and during our process we cubed twice, but the only roots we got were x=1 and x=3, both valid. So the solutions are x=1 and x=3.But let's also test another value just to be thorough. For example, let's take x=2:Left side: ∛2 + ∛(4 -3) = ∛2 +1 ≈1.26 +1≈2.26Right side: ∛[12(1)] = ∛12≈2.28These are not equal, so x=2 is not a solution, which is consistent with our roots found.Therefore, the solutions are x=1 and x=3.Final AnswerThe solutions are:(1) boxed{1} and boxed{-dfrac{a + 2b}{2a + b}};(2) boxed{1} and boxed{3}.</think>To solve the given equations, we proceed as follows:1. First Equation: [ (a + b)(a x + b)(a - b x) = (a^2 x - b^2)(a + b x) ] - Expand both sides and simplify: - Left side: ((a + b)(a x + b)(a - b x)) - Right side: ((a^2 x - b^2)(a + b x)) - After expanding and simplifying both sides, we find that the equation reduces to a quadratic equation in (x): [ (2a + b)x^2 - (a - b)x - (a + 2b) = 0 ] - Solving this quadratic equation using the quadratic formula, we get the solutions: [ x = 1 quad text{and} quad x = -frac{a + 2b}{2a + b} ] - Both solutions are verified to be valid.2. Second Equation: [ x^{frac{1}{3}} + (2 x - 3)^{frac{1}{3}} = {12(x - 1)}^{frac{1}{3}} ] - Cube both sides to eliminate the cube roots: [ left( x^{frac{1}{3}} + (2x - 3)^{frac{1}{3}} right)^3 = left( {12(x - 1)}^{frac{1}{3}} right)^3 ] - Simplify and cube again to eliminate the remaining cube root: [ 24x^3 - 60x^2 + 36x = (3x - 3)^3 ] - This results in the cubic equation: [ x^3 - 7x^2 + 15x - 9 = 0 ] - Factoring the cubic equation, we find the roots: [ x = 1 quad text{(double root)} quad text{and} quad x = 3 ] - Both solutions are verified to be valid. Final AnswerThe solutions are:(1) boxed{1} and boxed{-dfrac{a + 2b}{2a + b}};(2) boxed{1} and boxed{3}.]

🔑:(1) It is obvious that x=1 is a solution to the equation. Simplify this equation tobegin{array}{l}quad(a+b)left{a b+left(a^{2}-b^{2}right) x-a b x^{2}right} =a^{3} x-a b^{2}+a^{2} b x^{2}-b^{3} x . quad text { From the equation }left[a^{2} b+a b(a+b)right) x^{2} +left[a^{3}-i^{2}-(a+i)left(a^{2}-b^{2}right)right] x -left[a b^{2}+a b(a+b)right]=0end{array}we can obtain the product of the two roots asbegin{aligned}x_{1} cdot x_{2} & =-frac{a b^{2}+a b(a+b)}{a^{2} b+a b(a+b)} & =-frac{a+2 b}{2 a+b} .end{aligned}Thus, the second root can be easily found, which is-frac{a+2 b}{2 a+b} text {. }(2) Let c=a+b. Then,begin{aligned}c^{3} & =(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) & =a^{3}+b^{3}+3 a b c .end{aligned}Rearranging gives3 a b c=c^{3}-a^{3}-b^{3} .Therefore, the equivalent equation can be written asbegin{aligned}& 3 sqrt{12 x}(2 x-3)(x-1) = & 12(x-1)-x-(2 x-3)=9(x-1) .end{aligned}Cubing both sides yields12 x(2 x-3)(x-1)=27(x-1)^{3} .Thus,x-1=0 text { or } 4 x(2 x-3)=9(x-1)^{2} text {. }Therefore, x=1, x=3 are the roots of the equation.

🧠:[<think>Okay, let's see. I have this equation here: 2 times (x minus 1) equals x plus 1. Hmm, I need to solve for x. Alright, step by step. First, I should probably expand the left side to get rid of the parentheses. So, 2 multiplied by (x - 1). Let me do that. 2 times x is 2x, and 2 times -1 is -2. So, the left side becomes 2x - 2. The equation now looks like 2x - 2 = x + 1. Okay, that seems simpler. Now, I need to get all the x terms on one side and the constants on the other. Maybe subtract x from both sides to get rid of the x on the right. Let's try that.2x - 2 - x = x + 1 - x. Simplifying both sides, that would be (2x - x) - 2 = 1. Which is x - 2 = 1. Alright, now I need to solve for x. So, add 2 to both sides to get rid of the -2. x - 2 + 2 = 1 + 2. That simplifies to x = 3. Wait, is that right? Let me check my steps again to make sure I didn't make a mistake. Original equation: 2(x - 1) = x + 1. Expand left side: 2x - 2 = x + 1. Subtract x from both sides: 2x - 2 - x = x + 1 - x → x - 2 = 1. Add 2 to both sides: x = 3. Hmm, seems correct. Maybe plug the value back into the original equation to verify.Left side: 2(3 - 1) = 2(2) = 4. Right side: 3 + 1 = 4. Yep, both sides equal 4. So, x = 3 is the correct solution. Wait, just to make sure I didn't skip any steps or make any sign errors. Let's go through it once more. Starting over:Original equation: 2(x - 1) = x + 1.Left side: 2 times x is 2x, 2 times -1 is -2. So, 2x - 2.Set equal to right side: 2x - 2 = x + 1.Subtract x from both sides: 2x - 2 - x = 1 → x - 2 = 1.Add 2 to both sides: x = 3.Check substitution: 2(3 -1) = 2*2 = 4; 3 +1 = 4. Checks out. Alright, I think that's solid. No mistakes noticed here. So, the solution is x equals 3. I don't see any alternative methods needed here; the equation is straightforward. Maybe another way to look at it is by moving all terms to one side? Let's try that for thoroughness.Starting again: 2(x -1) - (x +1) = 0.Expand: 2x - 2 -x -1 = 0. Combine like terms: (2x -x) + (-2 -1) = 0 → x -3 = 0 → x = 3. Same result. Yep, definitely x=3.So, confident that's the answer. Not sure what else could go wrong here. Maybe if I distributed incorrectly, but I checked that. Or if I messed up signs when moving terms, but again, checked that. Alright, I think that's it.Final AnswerThe solution is boxed{3}.</think>To solve the equation (2(x - 1) = x + 1):1. Expand the left side: [ 2(x - 1) = 2x - 2 ]2. The equation becomes: [ 2x - 2 = x + 1 ]3. Subtract (x) from both sides: [ 2x - 2 - x = x + 1 - x implies x - 2 = 1 ]4. Add 2 to both sides: [ x - 2 + 2 = 1 + 2 implies x = 3 ]Verification:- Substitute (x = 3) back into the original equation: - Left side: (2(3 - 1) = 2 cdot 2 = 4) - Right side: (3 + 1 = 4)- Both sides are equal, confirming the solution is correct.Thus, the solution is (boxed{3}).,<think>Okay, so I need to solve the equation 2 times (x minus 1) equals x plus 1. Let me write that down first to make sure I have it right: 2(x - 1) = x + 1. Alright, now I need to figure out what value of x makes this equation true. Let me think about how to approach this.First, maybe I should start by expanding the left side of the equation. The left side is 2 multiplied by (x - 1). To expand that, I can use the distributive property, which means multiplying 2 by each term inside the parentheses. So, 2 times x is 2x, and 2 times -1 is -2. So after expanding, the left side becomes 2x - 2. That gives me the equation 2x - 2 = x + 1. Hmm, okay, that seems simpler. Now, I need to get all the x terms on one side and the constants on the other side. Let me subtract x from both sides to get rid of the x on the right. If I subtract x from both sides, the equation becomes 2x - 2 - x = x + 1 - x. Simplifying both sides, on the left, 2x minus x is x, so left with x - 2. On the right, x minus x cancels out, leaving just 1. So now the equation is x - 2 = 1.Now, I need to solve for x. To isolate x, I can add 2 to both sides. So, x - 2 + 2 = 1 + 2. Simplifying that, the left side becomes x, and the right side is 3. Therefore, x equals 3. Let me check if that's correct by plugging it back into the original equation.Original equation: 2(x - 1) = x + 1. Substitute x with 3: 2(3 - 1) = 3 + 1. Simplify inside the parentheses first: 3 - 1 is 2. Then, 2 times 2 is 4. On the right side, 3 + 1 is also 4. So both sides equal 4, which means x = 3 is indeed the correct solution. Wait, but let me go through the steps again to make sure I didn't skip anything or make a mistake. Starting with 2(x - 1) = x + 1. Distribute the 2: 2x - 2 = x + 1. Subtract x from both sides: x - 2 = 1. Add 2 to both sides: x = 3. Check the solution by substitution: left side is 2(3 - 1) = 4, right side is 3 + 1 = 4. Yep, that works. Is there another way to solve this equation? Maybe by moving all terms to one side first? Let's try that. Starting with 2(x - 1) = x + 1. Bring all terms to the left side by subtracting x and 1 from both sides: 2(x - 1) - x - 1 = 0. Let's expand that: 2x - 2 - x - 1 = 0. Combine like terms: (2x - x) + (-2 - 1) = 0 → x - 3 = 0. Then, x = 3. Same result. So that checks out too.Alternatively, maybe divide both sides by 2 first? Let's see. Original equation: 2(x - 1) = x + 1. Dividing both sides by 2: (x - 1) = (x + 1)/2. Then, multiply both sides by 2 to eliminate the denominator: 2(x - 1) = x + 1. Wait, that just brings us back to the original equation. So that approach doesn't really simplify things. Maybe that wasn't helpful.Another method: graphical interpretation. If we consider the left side as a function y = 2(x - 1) and the right side as y = x + 1, solving the equation is equivalent to finding the x-coordinate where these two lines intersect. Let's see. The first line, y = 2x - 2, has a slope of 2 and y-intercept at -2. The second line, y = x + 1, has a slope of 1 and y-intercept at 1. The intersection occurs where 2x - 2 = x + 1, which leads to the same equation as before. Solving that gives x = 3. Then plugging back in, y = 3 + 1 = 4, so the point of intersection is (3, 4). That makes sense.Alternatively, could there be no solution or infinite solutions? For linear equations, if the slopes are different, there is exactly one solution. If the slopes and y-intercepts are the same, infinite solutions; if same slope but different y-intercepts, no solution. Here, the slopes are 2 and 1, which are different, so exactly one solution. Which we found as x=3. So that's consistent.Alternatively, using algebra tiles or manipulatives. But since I can't visualize that here, maybe not necessary. But just thinking, if I had two groups of (x - 1) on the left and one x and one positive unit on the right. To balance, removing one x from both sides would leave me with x - 2 on the left and 1 on the right, then adding two units to both sides gives x = 3. Same process as before.Wait, maybe another check. If I take x = 3, then left side is 2*(3-1) = 2*2 = 4, right side is 3 + 1 = 4. Correct. What if I pick a different number, say x = 2. Left side: 2*(2 -1) = 2*1 = 2. Right side: 2 + 1 = 3. Not equal. So x=2 is not a solution. How about x=4? Left: 2*(4-1)=6, Right:4+1=5. Not equal. So only x=3 works.Let me check if I made any arithmetic mistakes. Let's go step by step again. Original equation: 2(x - 1) = x + 1. Distribute: 2x - 2 = x + 1. Subtract x: 2x - 2 - x = x + 1 - x → x - 2 = 1. Add 2: x - 2 + 2 = 1 + 2 → x = 3. All steps are correct. The operations are valid because when we subtract x or add 2, we do it to both sides, maintaining equality.Another way to look at it: the difference between the left and the right side. The left side is twice (x -1), the right is x +1. The difference between twice (x -1) and (x +1) is zero when x=3. Let's compute: 2*(3 -1) - (3 +1) = 2*2 -4 = 4 -4 =0. So indeed, the difference is zero, meaning they are equal.Alternatively, could set up a table of values for x and compute both sides:x | 2(x-1) | x+11 | 2(0)=0 | 22 | 2(1)=2 | 33 | 2(2)=4 |44 | 2(3)=6 |5So at x=3, both sides are 4. So again confirms that.Alternatively, thinking in terms of word problems. Suppose you have twice the difference of a number and one is equal to the number plus one. Find the number. Translating to equation: 2(x -1) =x +1. Solving gives x=3. Makes sense.Is there a way to represent this equation graphically? As I thought before, two lines intersecting at (3,4). Plotting y=2x -2 and y=x +1. The first line crosses the y-axis at (0,-2) and has a steeper slope. The second line crosses at (0,1) and has a gentler slope. They intersect at (3,4). So x=3 is the solution. Another check: using substitution step-by-step. Suppose x=3. Left side: 2*(3-1)=2*2=4. Right side:3+1=4. Equal. So correct. If x were something else, say 0: left=2*(-1)=-2, right=0+1=1. Not equal. x= -1: left=2*(-2)=-4, right= -1+1=0. Not equal. So x=3 is unique.Alternatively, solving for x by isolating variables. Let's try again:Start with 2(x -1) = x +1.Divide both sides by 2? Then: x -1 = (x +1)/2.Then, maybe add 1 to both sides: x = (x +1)/2 +1.Multiply both sides by 2: 2x = x +1 +2 → 2x =x +3.Subtract x: x=3. Same result. So that works too. Different path, same answer.Alternatively, start by expanding both sides if possible. But the right side is already simplified. So expanding the left gives 2x -2. So we can't expand further. Then collect like terms. Which we did.Alternatively, bringing all terms to the left: 2(x -1) -x -1 =0. Then 2x -2 -x -1=0 → x -3=0 →x=3. Same as before.So regardless of the method, we get x=3, which checks out in substitution.Therefore, confident that the solution is x=3.Final AnswerThe solution is boxed{3}.</think>To solve the equation (2(x - 1) = x + 1):1. Expand the left side using the distributive property: [ 2(x - 1) = 2x - 2 ] This transforms the equation into: [ 2x - 2 = x + 1 ]2. Subtract (x) from both sides to get all (x) terms on one side: [ 2x - 2 - x = x + 1 - x ] Simplifying both sides: [ x - 2 = 1 ]3. Add 2 to both sides to isolate (x): [ x - 2 + 2 = 1 + 2 ] Simplifying both sides: [ x = 3 ]4. Verify the solution by substituting (x = 3) back into the original equation: - Left side: (2(3 - 1) = 2 times 2 = 4) - Right side: (3 + 1 = 4)Both sides are equal, confirming that (x = 3) is correct. Final AnswerThe solution is (boxed{3}).]

🔑:To solve the equation 2(x-1)=x+1, we follow these steps:1. Expand the brackets: [2(x-1) = 2x - 2]So, the equation becomes:[2x - 2 = x + 1]2. Rearrange the terms to isolate x:Subtract x from both sides to get all x terms on one side:[2x - x = 1 + 2]3. Combine like terms:[x = 3]Therefore, the solution to the equation is boxed{x = 3}.

🧠:[<think>Okay, so I have this problem here with two functions, f(x) and g(x). Let me start by writing down what's given. f(x) is √2 times sin(ωx + φ), and g(x) is √2 times cos(ωx). The parameters are ω > 0 and φ is between 0 and π. Both functions have a minimum positive period of π. Also, one of the zero points of f(x) + g(x) is at -π/6. Then there are four options, A to D, and I need to figure out which ones are correct.First, let me recall that the period of a sine or cosine function is 2π divided by the coefficient of x inside the function. Since both f and g have a period of π, that should help me find ω. For f(x) = √2 sin(ωx + φ), the period is 2π / ω. The problem states that the minimum positive period is π, so 2π / ω = π. Solving for ω gives ω = 2. Let me check that. If ω were 2, then 2π / 2 = π, which is correct. Similarly, for g(x) = √2 cos(ωx), the period is also 2π / ω, so same result. So ω is indeed 2. That seems straightforward.So now I can update the functions:f(x) = √2 sin(2x + φ)g(x) = √2 cos(2x)Now, the next part is that a zero point of f(x) + g(x) is at -π/6. Let me write out f(x) + g(x):f(x) + g(x) = √2 sin(2x + φ) + √2 cos(2x)Factor out √2:= √2 [ sin(2x + φ) + cos(2x) ]We can use a trigonometric identity to combine these terms. Maybe using the sine addition formula or something. Let me see. Alternatively, perhaps express sin(2x + φ) as sin(2x)cosφ + cos(2x)sinφ. Let's try that.So:sin(2x + φ) = sin2x cosφ + cos2x sinφTherefore, f(x) + g(x) becomes:√2 [ sin2x cosφ + cos2x sinφ + cos2x ]= √2 [ sin2x cosφ + cos2x (sinφ + 1) ]Hmm, not sure if that's helpful. Maybe instead, let's try to write this as a single sine or cosine function. The expression is of the form A sinθ + B cosθ, which can be written as C sin(θ + δ) or C cos(θ + δ), where C = √(A² + B²). Let's try that.Given that:sin(2x + φ) + cos2x = sin2x cosφ + cos2x sinφ + cos2x= sin2x cosφ + cos2x (sinφ + 1)So, if we think of this as:A sin2x + B cos2xwhere A = cosφ and B = sinφ + 1Therefore, this can be written as:C sin(2x + δ)where C = √(A² + B²) = √(cos²φ + (sinφ + 1)^2 )Let me compute that:cos²φ + (sinφ + 1)^2 = cos²φ + sin²φ + 2 sinφ + 1But cos²φ + sin²φ = 1, so:1 + 2 sinφ + 1 = 2 + 2 sinφ = 2(1 + sinφ)Therefore, C = √[2(1 + sinφ)] = √2 √(1 + sinφ)Therefore, f(x) + g(x) = √2 * [ sin(2x + φ) + cos2x ] = √2 * [ A sin2x + B cos2x ] = √2 * C sin(2x + δ )Wait, but actually, since we already factored out √2, maybe I should consider the entire expression.Wait, let's step back. The original f(x) + g(x) is √2 sin(2x + φ) + √2 cos2x. So factor out √2:√2 [ sin(2x + φ) + cos2x ]Then, combining sin(2x + φ) + cos2x, which we can write as:sin(2x + φ) + cos2x = sin2x cosφ + cos2x sinφ + cos2x= sin2x cosφ + cos2x (sinφ + 1)So, this is of the form A sin2x + B cos2x where A = cosφ and B = sinφ + 1.So then, combining these terms into a single sine function:A sin2x + B cos2x = C sin(2x + δ)where C = √(A² + B²) = √[cos²φ + (sinφ + 1)^2]Which we calculated as √[2(1 + sinφ)].Then, the phase shift δ is given by tanδ = B / A = [sinφ + 1] / cosφTherefore, δ = arctan( [sinφ + 1] / cosφ )Alternatively, this can be written as a cosine function with a phase shift, but maybe that's more complicated.Alternatively, perhaps there is a better way to combine these terms. Let me think.Alternatively, use the identity for sum of sine and cosine. Let me recall that a sinθ + b cosθ = √(a² + b²) sin(θ + α), where α = arctan(b/a) or something like that. Wait, actually, the formula is:a sinθ + b cosθ = √(a² + b²) sin(θ + α), where α = arctan(b/a) if a ≠ 0. But depending on the signs, it might adjust.Alternatively, it's sometimes written as:a sinθ + b cosθ = √(a² + b²) cos(θ - δ), where δ = arctan(a/b). Hmm. Maybe it's better to use the sine addition formula.But in any case, the point is that the amplitude C is √(a² + b²). So here, since inside the brackets, we have:[ sin(2x + φ) + cos2x ] = sin2x cosφ + cos2x sinφ + cos2x = sin2x cosφ + cos2x (sinφ + 1)So A = cosφ and B = sinφ + 1. Therefore, the amplitude is √(cos²φ + (sinφ + 1)^2 ) = √(2 + 2 sinφ )So, then f(x) + g(x) = √2 * [ A sin2x + B cos2x ] = √2 * √(2 + 2 sinφ ) * sin(2x + δ )Wait, but actually, the total amplitude would be √2 multiplied by √(2 + 2 sinφ ). So the maximum value of f(x) + g(x) is √2 * √(2 + 2 sinφ ). Let me compute that.√2 * √(2 + 2 sinφ ) = √2 * √[2(1 + sinφ)] = √2 * √2 * √(1 + sinφ ) = 2√(1 + sinφ )So the maximum value is 2√(1 + sinφ ). Now, the question is, in option A, it says the maximum value is 2. So that would require that √(1 + sinφ ) = 1, which would mean 1 + sinφ = 1, so sinφ = 0. But φ is in [0, π). So sinφ = 0 implies φ = 0 or π. But φ is in [0, π), so φ = 0. But then φ can be 0? Wait, the problem says φ ∈ [0, π). So φ can be 0, but in that case, if φ = 0, then f(x) + g(x) becomes √2 sin2x + √2 cos2x. Let me compute the amplitude:√( (√2)^2 + (√2)^2 ) = √(2 + 2 ) = √4 = 2. So maximum value is 2, which would make option A correct. But wait, if φ is 0, then φ is allowed. But in our problem, we need to determine if the maximum value is 2 regardless of φ? Wait, no. Because φ is given, but we have to determine φ from the given zero point. So maybe φ is not zero. So maybe the maximum value isn't necessarily 2. Let's see.So we need to use the information that x = -π/6 is a zero of f(x) + g(x). So let's plug x = -π/6 into f(x) + g(x) and set it equal to zero.So f(-π/6) + g(-π/6) = 0.Compute f(-π/6):√2 sin(2*(-π/6) + φ ) = √2 sin(-π/3 + φ )Compute g(-π/6):√2 cos(2*(-π/6)) = √2 cos(-π/3) = √2 cos(π/3) = √2*(1/2) = √2/2Therefore:√2 sin(-π/3 + φ ) + √2/2 = 0Divide both sides by √2:sin(-π/3 + φ ) + 1/2 = 0So sin(-π/3 + φ ) = -1/2Therefore:-π/3 + φ = arcsin(-1/2)But arcsin(-1/2) is equal to -π/6 or 7π/6, but considering the range of arcsin, which is [-π/2, π/2], so arcsin(-1/2) = -π/6. However, sine is periodic, so general solution is:-π/3 + φ = -π/6 + 2πk or π + π/6 + 2πk, where k is integer.Therefore,First solution:-π/3 + φ = -π/6 + 2πk=> φ = -π/6 + π/3 + 2πk = π/6 + 2πkSecond solution:-π/3 + φ = π + π/6 + 2πk = 7π/6 + 2πk=> φ = 7π/6 + π/3 + 2πk = 7π/6 + 2π/6 = 9π/6 = 3π/2 + 2πkBut φ is in [0, π), so we have to find φ in this interval.First solution:φ = π/6 + 2πkSince φ ∈ [0, π), the possible values are for k=0: φ=π/6; k=1: φ=π/6 + 2π, which is outside the interval. So only φ=π/6 is possible.Second solution:φ = 3π/2 + 2πk. But 3π/2 is 270 degrees, which is outside [0, π). So there's no solution here. So only φ=π/6 is valid.Therefore, φ=π/6.So now, we know φ is π/6. Therefore, the functions become:f(x) = √2 sin(2x + π/6)g(x) = √2 cos(2x)Now, let's check the maximum value of f(x) + g(x). Earlier, we had that the amplitude is 2√(1 + sinφ ). Since φ=π/6, sinφ = sin(π/6) = 1/2. Therefore, the amplitude is 2√(1 + 1/2 ) = 2√(3/2 ) = 2*(√6)/2 = √6 ≈ 2.449. Wait, that's different from my previous thought when I thought if φ=0, maximum value is 2. But since φ is π/6, the maximum value is √6. Therefore, option A says the maximum value is 2, which would be incorrect. Wait, but maybe my previous calculation was wrong.Wait, hold on. Let me re-examine. The original f(x) + g(x) is √2 sin(2x + φ) + √2 cos(2x). When I combine these terms, let's do it again with φ=π/6.So f(x) + g(x) = √2 sin(2x + π/6) + √2 cos(2x). Let's compute this.First, expand sin(2x + π/6):sin(2x)cos(π/6) + cos(2x)sin(π/6) = sin(2x)*(√3/2) + cos(2x)*(1/2)Therefore, f(x) + g(x) becomes:√2 [ (√3/2 sin2x + 1/2 cos2x ) + cos2x ]= √2 [ (√3/2 sin2x + 1/2 cos2x + cos2x ) ]= √2 [ √3/2 sin2x + (1/2 + 1) cos2x ]= √2 [ (√3/2) sin2x + (3/2) cos2x ]Factor out 1/2:= √2 * (1/2) [ √3 sin2x + 3 cos2x ]= (√2 / 2) [ √3 sin2x + 3 cos2x ]Alternatively, let's compute the amplitude:The coefficients are √3 and 3, so amplitude is √( (√3)^2 + 3^2 ) = √(3 + 9 ) = √12 = 2√3. Then multiplied by √2 / 2:Total amplitude = (√2 / 2) * 2√3 = √2 * √3 = √6 ≈ 2.449. So the maximum value is √6. Therefore, option A says the maximum is 2, which is incorrect. So A is wrong.Hmm, so maybe I made a mistake in the earlier approach. Wait, alternatively, let's compute f(x) + g(x) when φ=π/6. Let me compute it as:f(x) + g(x) = √2 sin(2x + π/6) + √2 cos(2x). Let's try to combine these into a single trigonometric function.We can write this as √2 [ sin(2x + π/6) + cos2x ]Let me use the identity sin(A) + cos(B). Hmm, but they have the same argument here, except for the phase. Let me adjust.Let me write cos2x as sin(2x + π/2), since cosθ = sin(θ + π/2). Therefore:sin(2x + π/6) + sin(2x + π/2)Using the identity sinA + sinB = 2 sin( (A+B)/2 ) cos( (A-B)/2 )So A = 2x + π/6, B = 2x + π/2Compute (A+B)/2 = (2x + π/6 + 2x + π/2)/2 = (4x + 2π/3)/2 = 2x + π/3Compute (A - B)/2 = (2x + π/6 - 2x - π/2)/2 = (-π/3)/2 = -π/6Therefore:sin(2x + π/6) + sin(2x + π/2) = 2 sin(2x + π/3) cos(-π/6) = 2 sin(2x + π/3) cos(π/6)Since cos(-π/6) = cos(π/6) = √3/2Therefore:= 2 sin(2x + π/3) * (√3/2) = √3 sin(2x + π/3)Therefore, f(x) + g(x) = √2 * √3 sin(2x + π/3) = √6 sin(2x + π/3)Therefore, the maximum value is √6, which is approximately 2.449, not 2. Therefore, option A is incorrect. So A is wrong.Moving on to option B: The graph of f(x) + g(x) is symmetric about the point (-2π/3, 0). Hmm, symmetry about a point means that if you rotate the graph 180 degrees around that point, it remains unchanged. In other words, for any point (a, b) on the graph, the point (2h - a, 2k - b) is also on the graph, where (h, k) is the center of symmetry. Here, the center is (-2π/3, 0), so for any x, f(-4π/3 - x) + g(-4π/3 - x) = - [f(x) + g(x)]Wait, maybe more carefully: If the function is symmetric about the point (h, k), then f(2h - x) + g(2h - x) = 2k - (f(x) + g(x)). Here, k=0, so f(2h - x) + g(2h - x) = - (f(x) + g(x)).Given that h = -2π/3, so 2h = -4π/3. Therefore, check if f(-4π/3 - x) + g(-4π/3 - x) = - [f(x) + g(x)]Alternatively, since we have f(x) + g(x) = √6 sin(2x + π/3), let's check if this function is symmetric about the point (-2π/3, 0).A function is symmetric about the point (a, b) if f(a + t) + f(a - t) = 2b for all t. But since our function is sinusoidal, maybe there's another way. Alternatively, since the function is √6 sin(2x + π/3), let's find its zeros and critical points.But perhaps a better approach is to check if for the point (-2π/3, 0), if f(-2π/3 + t) + g(-2π/3 + t) = - [f(-2π/3 - t) + g(-2π/3 - t) ]Let me compute f(-2π/3 - t) + g(-2π/3 - t):= √6 sin(2*(-2π/3 - t) + π/3) = √6 sin(-4π/3 - 2t + π/3) = √6 sin(-π - 2t) = √6 sin(-π -2t) = √6 [-sin(π + 2t)] = √6 [-(-sin2t)] = √6 sin2tWait, sin(-π -2t) = -sin(π + 2t) = -(-sin2t) = sin2t. Wait, hold on:sin(-π -2t) = sin(- (π + 2t)) = -sin(π + 2t) = -(-sin2t) = sin2tTherefore, f(-2π/3 -t) + g(-2π/3 -t) = √6 sin2tOn the other hand, - [f(-2π/3 + t) + g(-2π/3 + t) ] = -√6 sin(2*(-2π/3 + t) + π/3 )= -√6 sin(-4π/3 + 2t + π/3 ) = -√6 sin(-π + 2t ) = -√6 [ -sin(π - 2t ) ] = -√6 [ -sin2t ] = √6 sin2tWait, sin(-π + 2t) = -sin(π - 2t) = -sin2t, since sin(π - θ) = sinθ, but here θ = 2t, so sin(π - 2t) = sin2t if π - 2t is in the right quadrant? Wait, sin(π - θ) = sinθ, so sin(π - 2t) = sin2t. Therefore:sin(-π + 2t) = sin(- (π - 2t)) = -sin(π - 2t) = -sin2tTherefore:-√6 sin(-π + 2t ) = -√6 (-sin2t ) = √6 sin2tTherefore, indeed:f(-2π/3 -t) + g(-2π/3 -t) = √6 sin2tand- [f(-2π/3 + t) + g(-2π/3 + t) ] = √6 sin2tTherefore,f(-2π/3 -t) + g(-2π/3 -t) = - [f(-2π/3 + t) + g(-2π/3 + t) ]Which implies that the function is symmetric about the point (-2π/3, 0). Therefore, option B is correct.Wait, let me check again. If I take a point t units to the right of -2π/3, which is (-2π/3 + t, y), then the corresponding point on the left should be (-2π/3 - t, -y). So if f(-2π/3 -t) + g(-2π/3 -t) = - [f(-2π/3 + t) + g(-2π/3 + t) ], then that point (-2π/3 - t, y) would correspond to (-2π/3 + t, -y), which is the reflection over (-2π/3, 0). Therefore, yes, this symmetry holds. Therefore, B is correct.Moving on to option C: Both f(x) and g(x) are monotonically increasing on [2π/3, 5π/6].First, let's recall the derivatives. Since f(x) = √2 sin(2x + π/6), its derivative is f’(x) = √2 * 2 cos(2x + π/6) = 2√2 cos(2x + π/6)Similarly, g(x) = √2 cos(2x), derivative is g’(x) = -2√2 sin(2x)We need to check if both f’(x) and g’(x) are positive on [2π/3, 5π/6].First, let's compute the interval in terms of 2x:For x in [2π/3, 5π/6], then 2x is in [4π/3, 5π/3]First, check f’(x) = 2√2 cos(2x + π/6). So 2x + π/6 is in [4π/3 + π/6, 5π/3 + π/6] = [ (8π/6 + π/6 ), (10π/6 + π/6 ) ] = [9π/6, 11π/6] = [3π/2, 11π/6]3π/2 is 270 degrees, 11π/6 is 330 degrees. So cosine of angles between 270 and 330 degrees. Cosine is positive in the fourth quadrant (between 270 and 360), which is 3π/2 to 2π. But 11π/6 is 330 degrees, which is still in the fourth quadrant. So cos(θ) is positive in 270 < θ < 360 degrees, which is 3π/2 < θ < 2π. However, 3π/2 to 11π/6 is 3π/2 (270) to 11π/6 (330). So from 270 to 330 degrees. Cosine is positive from 270 to 360, which includes 270 to 330. Wait, but cos(3π/2) = 0, cos(11π/6) = cos(330°) = √3/2. So on the interval [3π/2, 11π/6], cosine increases from 0 to √3/2. Therefore, cos(theta) is non-negative here, so f’(x) = 2√2 cos(theta) is non-negative. Therefore, f(x) is non-decreasing on [2π/3, 5π/6]. But is it monotonically increasing? Well, since cos(theta) is strictly positive in (3π/2, 11π/6) except at theta=3π/2 where it's zero. But 2x + π/6 ranges from 3π/2 to 11π/6. So theta starts at 3π/2 (where cos(theta)=0) and goes to 11π/6 (cos(theta)=√3/2). Therefore, f’(x) starts at 0 and increases to 2√2*(√3/2) = √6. So the derivative is non-decreasing, but at the left endpoint x=2π/3, derivative is zero. Therefore, f(x) is increasing on (2π/3, 5π/6], but at x=2π/3, the derivative is zero. So technically, if the interval includes x=2π/3, then f(x) is not strictly increasing on the entire interval [2π/3, 5π/6], because at the left endpoint the derivative is zero. But maybe the problem allows for non-decreasing? Wait, the question says "monotonically increasing". Generally, that means strictly increasing, but sometimes it can mean non-decreasing. However, in calculus, "monotonically increasing" allows for intervals where the derivative is zero. Wait, the formal definition is: a function is monotonically increasing on an interval if for any x1 < x2 in the interval, f(x1) ≤ f(x2). So in that case, even if the derivative is zero at some points, it's still considered monotonically increasing. However, if the derivative is positive almost everywhere except at isolated points, it's still non-decreasing. So in our case, f’(x) is non-negative on [2π/3, 5π/6], with f’(x) = 0 only at x=2π/3, and positive elsewhere. Therefore, f(x) is monotonically increasing (non-decreasing) on the interval.Similarly, check g’(x) = -2√2 sin(2x). For x in [2π/3, 5π/6], 2x is in [4π/3, 5π/3]. So sin(2x) in this interval: 4π/3 is 240 degrees, 5π/3 is 300 degrees. So sine is negative in both quadrants (third and fourth). Therefore, sin(2x) is negative, so g’(x) = -2√2 sin(2x) = -2√2*(negative) = positive. Therefore, g’(x) is positive throughout the interval. Therefore, g(x) is monotonically increasing on [2π/3, 5π/6]. Therefore, both derivatives are non-negative on the interval. For f(x), the derivative is zero at the left endpoint but positive elsewhere, and for g(x), the derivative is positive throughout. So both functions are monotonically increasing (non-decreasing) on [2π/3, 5π/6]. Therefore, option C is correct.Now, option D: Shifting the graph of f(x) to the left by π/6 units will result in the graph of g(x).So shifting f(x) left by π/6 means replacing x with x + π/6 in f(x). Let's compute:f(x + π/6) = √2 sin[2(x + π/6) + π/6] = √2 sin(2x + π/3 + π/6) = √2 sin(2x + π/2)But sin(2x + π/2) = cos(2x), since sin(θ + π/2) = cosθ. Therefore, f(x + π/6) = √2 cos(2x) = g(x). Therefore, shifting f(x) to the left by π/6 units gives g(x). So option D is correct.Wait, but the question says shifting to the left by π/6, which is indeed replacing x with x + π/6. So yes, that would be correct.Wait, but wait, in the problem statement, f(x) is √2 sin(2x + φ), and we found φ = π/6. So f(x) is √2 sin(2x + π/6). Shifting left by π/6 gives √2 sin(2(x + π/6) + π/6) = √2 sin(2x + π/3 + π/6) = √2 sin(2x + π/2) = √2 cos(2x) = g(x). Therefore, D is correct.Wait, but hold on. Shifting to the left by π/6 is equivalent to replacing x with x + π/6. So yes, as we saw, this results in g(x). Therefore, D is correct.But wait, let me confirm once more. The original f(x) is √2 sin(2x + π/6). If we shift it left by π/6, then the new function is f(x + π/6) = √2 sin(2(x + π/6) + π/6) = √2 sin(2x + π/3 + π/6) = √2 sin(2x + π/2). Since sin(2x + π/2) = cos(2x), because sin(θ + π/2) = cosθ. Therefore, f(x + π/6) = √2 cos2x = g(x). Therefore, yes, shifting f(x) left by π/6 gives g(x). So D is correct.But wait, let me check the direction. Shifting left is adding to x, so replacing x with x + h shifts the graph left by h. So shifting left by π/6 would indeed be x → x + π/6. Therefore, D is correct.But wait, but in the problem statement, option D says: "Shifting the graph of f(x) to the left by π/6 units will result in the graph of g(x)". So according to our calculation, that's correct. Therefore, D is correct.Wait, but in the options, A was wrong, B correct, C correct, D correct. But let me check again.Wait, no. Wait, in the problem statement, the options are:A: The maximum value of f(x)+g(x) is 2B: The graph of f(x)+g(x) is symmetric about the point (−2π/3,0)C: Both f(x) and g(x) are monotonically increasing on [2π/3,5π/6]D: Shifting the graph of f(x) to the left by π/6 units will result in the graph of g(x)From our analysis:A: Wrong, maximum is √6B: CorrectC: CorrectD: CorrectBut let me verify once again for C. For the interval [2π/3, 5π/6], which is approximately [2.094, 2.618] in radians.For f(x) = √2 sin(2x + π/6). Let's check the derivative f’(x) = 2√2 cos(2x + π/6). At x=2π/3:2x + π/6 = 4π/3 + π/6 = 9π/6 = 3π/2. Cos(3π/2) = 0. At x=5π/6:2x + π/6 = 5π/3 + π/6 = 11π/6. Cos(11π/6) = cos(-π/6) = √3/2. So derivative starts at 0 and increases to √3/2 * 2√2 = √6. So derivative is increasing from 0 to √6. Hence, f’(x) ≥ 0 on [2π/3, 5π/6]. Therefore, f(x) is non-decreasing. So in calculus terms, that is monotonically increasing.For g(x) = √2 cos(2x), derivative is g’(x) = -2√2 sin(2x). For x in [2π/3,5π/6], 2x is in [4π/3,5π/3]. Sin(4π/3) = sin(π + π/3) = -sin(π/3) = -√3/2. Sin(5π/3) = sin(2π - π/3) = -sin(π/3) = -√3/2. Therefore, sin(2x) is negative throughout the interval. Therefore, g’(x) = -2√2 sin(2x) = positive. Hence, g’(x) > 0 on the entire interval. Therefore, g(x) is strictly increasing on [2π/3,5π/6].So f(x) is non-decreasing (since derivative starts at 0 and becomes positive), and g(x) is strictly increasing. Therefore, both are monotonically increasing (non-decreasing) on that interval. So C is correct.Therefore, options B, C, D are correct, and A is incorrect.But wait, the answer choices might be multiple, but the original problem didn't specify how many options are correct. The user just provided the question and says "then ()" with options. In Chinese exams, often multiple options can be correct. So according to my analysis, B, C, D are correct. But let me check once again.Wait, the problem says "both of which have a minimum positive period of π"—so we already derived ω=2.Then, f(x)+g(x) has a zero at -π/6, which led us to φ=π/6. Then:A: maximum value is 2. But we found it's √6, so A is wrong.B: symmetry about (-2π/3,0). Verified correct.C: both f and g are increasing on [2π/3,5π/6]. Verified correct.D: shifting f left by π/6 gives g. Verified correct.Therefore, options B, C, D are correct. But let me check the original problem again to ensure I didn't misread.Problem statement:Given functions f(x)=√2 sin(ω x + φ), g(x)=√2 cos(ω x), ω >0, φ ∈ [0,π), both of which have a minimum positive period of π, and a zero point of f(x)+g(x) is -π/6, then ()Options A to D as before.So according to this, B, C, D are correct. However, let me check option D once again. Shifting left by π/6 gives g(x). We saw that f(x + π/6) = g(x). So yes, shifting left by π/6 units. Therefore, D is correct.Therefore, the correct options are B, C, D.But wait, in the original problem, it's possible that the answer is only some of them. Let me check if the options are possibly different.But according to the user's original problem, the options are labeled A, B, C, D. The user hasn't specified if multiple can be correct or only one. In Chinese multiple-choice questions, often they have multiple correct options, and you have to select all that apply, but sometimes they might have only one. But in the absence of specific instructions, but given the way the problem is written, it's possible that multiple are correct.But let me check once again.A: Maximum is 2. As we found, it's √6. So A is wrong.B: Symmetric about (-2π/3, 0). Yes, as we verified.C: Both functions are increasing on [2π/3,5π/6]. Yes.D: Shifting f left by π/6 gives g. Yes.So correct answers are B, C, D.But the original problem is presented as a multiple-choice question where each option is a statement, possibly with multiple correct. However, the way the user presents the problem: "then ()" with options, which is common in Chinese tests where multiple options may be correct. The user hasn't specified if only one is correct, so we should check all.But according to the analysis, B, C, D are correct. However, let me check once more for option C.For the interval [2π/3, 5π/6]:For f(x): derivative starts at 0 and increases to √6. Therefore, it is non-decreasing. So in the context of the problem, since it says "monotonically increasing", which in mathematical terms means non-decreasing, so yes.For g(x): derivative is positive throughout, so strictly increasing. Therefore, both are monotonically increasing. So C is correct.Therefore, the answer should be B, C, D. But let me check if the original options are labeled as (A) (B) (C) (D), and the user might expect the answer to be the letters inside boxed{}.But since the user instruction says "put your final answer within boxed{}", and in Chinese exams, multiple answers are boxed like boxed{ABC} or similar. However, given that the user hasn't specified the format, but in the initial problem statement, the options are labeled A, B, C, D. If multiple are correct, the answer would be something like boxed{BCD}. But let me confirm once more.Yes, based on the analysis, options B, C, D are correct. So the final answer should be boxed{BCD}.Final Answerboxed{BCD}</think>Given functions ( f(x) = sqrt{2} sin(omega x + varphi) ) and ( g(x) = sqrt{2} cos(omega x) ) with (omega > 0) and (varphi in [0, pi)), both having a minimum positive period of (pi). A zero point of ( f(x) + g(x) ) is ( -frac{pi}{6} ).1. Determine (omega): - The period of both functions is (pi), so (omega = 2).2. Determine (varphi): - Given ( fleft(-frac{pi}{6}right) + gleft(-frac{pi}{6}right) = 0 ): - ( sqrt{2} sinleft(-frac{pi}{3} + varphiright) + sqrt{2} cosleft(-frac{pi}{3}right) = 0 ) - Solving, we find (varphi = frac{pi}{6}).3. Analyze ( f(x) + g(x) ): - Combine ( f(x) = sqrt{2} sin(2x + frac{pi}{6}) ) and ( g(x) = sqrt{2} cos(2x) ): - ( f(x) + g(x) = sqrt{6} sinleft(2x + frac{pi}{3}right) ) - Maximum value is (sqrt{6}), so A is incorrect.4. Symmetry about (-frac{2pi}{3}, 0): - Verify symmetry by checking ( f(-2pi/3 - t) + g(-2pi/3 - t) = -[f(-2pi/3 + t) + g(-2pi/3 + t)] ). Confirmed, so B is correct.5. Monotonicity on ([2pi/3, 5pi/6]): - Derivatives ( f'(x) ) and ( g'(x) ) are non-negative and positive respectively, so C is correct.6. Shift ( f(x) ) left by (frac{pi}{6}): - ( f(x + frac{pi}{6}) = sqrt{2} cos(2x) = g(x) ), so D is correct.Final answer: (boxed{BCD}),<think>Okay, let's try to tackle this problem step by step. First, let me read through the question again to make sure I understand all the parts. So, we have two functions: f(x) = √2 sin(ωx + φ) and g(x) = √2 cos(ωx). Both have a minimum positive period of π, and ω is greater than 0. The phase shift φ is between 0 and π. Also, there's a zero point of the sum f(x) + g(x) at x = -π/6. Then there are four statements (A to D) we need to check for correctness.Alright, starting with the given information. Both functions have a period of π. The period of a sine or cosine function is 2π divided by the coefficient of x inside the function, which here is ω. So for both f(x) and g(x), the period is 2π/ω. Since the minimum positive period is π, that means 2π/ω = π. Solving for ω, we get ω = 2. Let me confirm that: if ω = 2, then the period is 2π/2 = π, which matches. So ω is 2. Got that down.So now, f(x) = √2 sin(2x + φ) and g(x) = √2 cos(2x). The next piece of information is that x = -π/6 is a zero point of f(x) + g(x). So, substituting x = -π/6 into f(x) + g(x) gives 0. Let's write that out:f(-π/6) + g(-π/6) = 0Substituting the functions:√2 sin(2*(-π/6) + φ) + √2 cos(2*(-π/6)) = 0Simplify the arguments:First term: sin(-π/3 + φ)Second term: cos(-π/3)Since cosine is even, cos(-π/3) = cos(π/3) = 1/2.So the equation becomes:√2 [sin(φ - π/3) + 1/2] = 0Divide both sides by √2:sin(φ - π/3) + 1/2 = 0Therefore:sin(φ - π/3) = -1/2Now, we need to solve for φ given that φ ∈ [0, π). Let's set θ = φ - π/3. Then sinθ = -1/2. The solutions for θ in the range [ -π/3, π - π/3 ) = [ -π/3, 2π/3 ) since φ is between 0 and π.The general solution for sinθ = -1/2 is θ = 7π/6 + 2πk or θ = 11π/6 + 2πk, but considering θ must be in [ -π/3, 2π/3 ). Let's see. Let me convert -1/2 to angles. The sine is -1/2 at 7π/6 and 11π/6 in the unit circle. But θ must be between -π/3 (≈ -1.047) and 2π/3 (≈ 2.094). 7π/6 is ≈ 3.665, which is outside, 11π/6 ≈ 5.759, also outside. So perhaps we need to find an angle within the range by subtracting 2π?Wait, but θ = φ - π/3, and φ is between 0 and π, so θ is between -π/3 and 2π/3. So we need to find θ in that interval where sinθ = -1/2.Let's think. The sine is -1/2 at 7π/6 and 11π/6. If we subtract 2π from 7π/6, we get 7π/6 - 2π = -5π/6 ≈ -2.618, which is less than -π/3, so not in the interval. Alternatively, maybe adding or subtracting π? Let's think. Alternatively, maybe looking for reference angles. The reference angle for sinθ = -1/2 is π/6. So in the third and fourth quadrants. But θ is between -π/3 and 2π/3. So negative angles go from -π/3 to 0, and positive from 0 to 2π/3. So in the negative part, θ = -π/6 (since sin(-π/6) = -1/2). Let's check that. If θ = -π/6, then sin(-π/6) = -1/2. So θ = -π/6 is in the interval [ -π/3, 2π/3 ). So that's a solution. Alternatively, θ = 7π/6 - 2π = 7π/6 - 12π/6 = -5π/6, which is too low. So the only solution in the interval is θ = -π/6. Therefore:θ = φ - π/3 = -π/6 => φ = -π/6 + π/3 = π/6.So φ = π/6. Let me check that. φ is in [0, π), and π/6 is within that interval. Good.So now, we know that φ = π/6 and ω = 2. So the functions are:f(x) = √2 sin(2x + π/6)g(x) = √2 cos(2x)Now, let's check the statements A to D.A: The maximum value of f(x) + g(x) is 2First, let's compute f(x) + g(x). Let's combine the two functions:f(x) + g(x) = √2 sin(2x + π/6) + √2 cos(2x)We can factor out √2:√2 [ sin(2x + π/6) + cos(2x) ]Let me try to simplify the expression inside the brackets. Let's use the sine addition formula on sin(2x + π/6):sin(2x + π/6) = sin(2x)cos(π/6) + cos(2x)sin(π/6) = sin(2x)(√3/2) + cos(2x)(1/2)Therefore:sin(2x + π/6) + cos(2x) = (√3/2) sin(2x) + (1/2) cos(2x) + cos(2x) = (√3/2) sin(2x) + (3/2) cos(2x)So f(x) + g(x) = √2 [ (√3/2 sin2x + 3/2 cos2x ) ]Alternatively, we can write this as √2 times a linear combination of sin2x and cos2x. Maybe we can express this as a single sine (or cosine) function with phase shift. Let's see.The general form is A sin2x + B cos2x. Here, A = √3/2, B = 3/2. The amplitude is sqrt(A² + B²). Let's compute that:sqrt( ( (√3/2)^2 + (3/2)^2 ) ) = sqrt( (3/4 + 9/4 ) ) = sqrt(12/4) = sqrt(3) ≈ 1.732So the amplitude is sqrt(3), so the maximum value of the expression inside the brackets is sqrt(3). Then multiplying by √2, the maximum of f(x) + g(x) is √2 * sqrt(3) = sqrt(6) ≈ 2.449. But option A says the maximum is 2. So is that correct?Wait, wait. Maybe I made a mistake in the calculation. Let me check again.Wait, f(x) + g(x) = √2 [ sin(2x + π/6) + cos2x ]Alternatively, maybe there's a better way to combine them. Let me try to add sin(2x + π/6) + cos(2x). Let's use the identity that sin(a) + cos(b) can be combined, but maybe another approach.Alternatively, let's write both terms as sin and cos and see:sin(2x + π/6) + cos2x = sin2x cos(π/6) + cos2x sin(π/6) + cos2x = sin2x*(√3/2) + cos2x*(1/2) + cos2x = (√3/2) sin2x + (3/2) cos2x.Yes, that's correct. Then, the amplitude is sqrt( (√3/2)^2 + (3/2)^2 ) = sqrt( 3/4 + 9/4 ) = sqrt(12/4) = sqrt(3). So the maximum of the bracketed term is sqrt(3), so the maximum of f(x) + g(x) is √2 * sqrt(3) = sqrt(6). So sqrt(6) ≈ 2.449, which is greater than 2. Therefore, statement A is incorrect. So A is false. Hmm. But let's make sure.Wait, perhaps there's another way to combine the functions. Let me think. Alternatively, let's consider f(x) + g(x) = √2 sin(2x + π/6) + √2 cos2x. Let's factor out √2:√2 [ sin(2x + π/6) + cos2x ]Alternatively, maybe we can write this as a single sinusoidal function. Let's set:sin(2x + π/6) + cos2x = C sin(2x + θ) + D cos(2x + θ)? Wait, not sure. Alternatively, use the formula for sum of sinusoids.Alternatively, let's write sin(2x + π/6) as sin2x cos(π/6) + cos2x sin(π/6) = (√3/2) sin2x + (1/2) cos2x, as before. Then adding cos2x gives (√3/2) sin2x + (3/2) cos2x. Then, this is equal to sqrt( ( (√3/2)^2 + (3/2)^2 ) ) * sin(2x + φ), where φ = arctan( (3/2) / (√3/2) ) = arctan(3/√3) = arctan(√3) = π/3. So:(√3/2 sin2x + 3/2 cos2x) = sqrt( ( (√3/2)^2 + (3/2)^2 )) sin(2x + π/3 ) = sqrt(3/4 + 9/4) sin(2x + π/3) ) = sqrt(12/4) sin(2x + π/3) = sqrt(3) sin(2x + π/3)Therefore, f(x) + g(x) = √2 * sqrt(3) sin(2x + π/3) = sqrt(6) sin(2x + π/3)So the amplitude is sqrt(6), so the maximum value is sqrt(6) ≈ 2.449, which is greater than 2, so statement A is wrong. Therefore, A is false. So answer A is incorrect.B: The graph of f(x) + g(x) is symmetric about the point (-2π/3, 0)Hmm, symmetry about a point. To check if a function is symmetric about a point (a, b), we need to verify that for any point (x, y) on the graph, the point (2a - x, 2b - y) is also on the graph. Since it's symmetry about (-2π/3, 0), we need to check if for any x, f(2*(-2π/3) - x) + g(2*(-2π/3) - x) = - [f(x) + g(x)]Wait, more precisely, if the function h(x) = f(x) + g(x) is symmetric about the point (a, 0), then h(2a - x) = -h(x). So here, a = -2π/3. So we need to check if h(-4π/3 - x) = -h(x). Let's compute h(x):h(x) = sqrt(6) sin(2x + π/3)So let's compute h(-4π/3 - x):sqrt(6) sin[ 2*(-4π/3 - x) + π/3 ] = sqrt(6) sin( -8π/3 - 2x + π/3 ) = sqrt(6) sin( -7π/3 - 2x )But sin(-7π/3 - 2x) = sin( - (7π/3 + 2x ) ) = -sin(7π/3 + 2x )But 7π/3 is equivalent to 7π/3 - 2π = 7π/3 - 6π/3 = π/3. So sin(7π/3 + 2x) = sin(π/3 + 2x). Therefore:sqrt(6) * [ -sin(π/3 + 2x) ] = -sqrt(6) sin(π/3 + 2x )Compare to -h(x):-h(x) = -sqrt(6) sin(2x + π/3 )So we have h(-4π/3 - x) = -sqrt(6) sin(π/3 + 2x ) = -h(x). Therefore, yes, h(-4π/3 - x) = -h(x), which implies that the function is symmetric about the point (-2π/3, 0). Because the midpoint between x and -4π/3 - x is (-2π/3 - x/2) ??? Wait, maybe another way: For any x, the point (x, h(x)) has a corresponding point (2*(-2π/3) - x, -h(x)) which is (-4π/3 - x, -h(x)). So if we take h(-4π/3 - x) = -h(x), that means that the graph is symmetric about (-2π/3, 0). Therefore, statement B is correct. So B is true.C: Both f(x) and g(x) are monotonically increasing on [2π/3, 5π/6]We need to check the monotonicity of both functions on that interval. Let's first recall that f(x) = √2 sin(2x + π/6) and g(x) = √2 cos(2x). The derivative of f(x) is f’(x) = √2 * 2 cos(2x + π/6) = 2√2 cos(2x + π/6). Similarly, the derivative of g(x) is g’(x) = -√2 * 2 sin(2x) = -2√2 sin(2x). We need to check if both derivatives are positive on [2π/3, 5π/6], which would mean the functions are increasing.First, let's compute the interval for 2x + π/6 in f’(x):When x is in [2π/3, 5π/6], 2x is in [4π/3, 5π/3]. Then 2x + π/6 is in [4π/3 + π/6, 5π/3 + π/6] = [9π/6, 11π/6] = [3π/2, 11π/6]. So 2x + π/6 goes from 3π/2 to 11π/6. The cosine of angles in this range: 3π/2 is 270 degrees, where cosine is 0. From 3π/2 to 11π/6 (which is 330 degrees). So cosine is positive in the fourth quadrant, between 270 and 360 degrees (i.e., between 3π/2 to 2π). But 11π/6 is less than 2π (which is 12π/6), so from 3π/2 to 11π/6, cosine is positive. Wait, wait. 3π/2 is 270 degrees, and 11π/6 is 330 degrees. So in this interval, cosine goes from 0 at 270 degrees to positive (cos(330°) = √3/2). So cos(θ) is positive in (3π/2, 11π/6). Therefore, f’(x) = 2√2 cos(2x + π/6) is positive in [2π/3, 5π/6], so f(x) is increasing on this interval.Now for g’(x) = -2√2 sin(2x). Let's check when x is in [2π/3, 5π/6], 2x is in [4π/3, 5π/3]. So sin(2x) in this interval. 4π/3 is 240 degrees, 5π/3 is 300 degrees. So sine is negative in both third and fourth quadrants. So sin(2x) is negative in [4π/3, 5π/3]. Therefore, g’(x) = -2√2 sin(2x) = -2√2*(negative) = positive. Therefore, g’(x) is positive on [2π/3, 5π/6], so g(x) is increasing on that interval.Therefore, both f and g are monotonically increasing on [2π/3, 5π/6]. So statement C is correct. So C is true.D: Shifting the graph of f(x) to the left by π/6 units will result in the graph of g(x)Let's check. Shifting f(x) left by π/6 means replacing x with x + π/6 in f(x). So:f(x + π/6) = √2 sin(2(x + π/6) + π/6) = √2 sin(2x + π/3 + π/6) = √2 sin(2x + π/2)But sin(2x + π/2) = cos(2x), since sin(θ + π/2) = cosθ. Therefore:f(x + π/6) = √2 cos(2x) = g(x)So shifting f(x) to the left by π/6 units gives g(x). But shifting left by π/6 would mean replacing x with x + π/6, which is a shift to the left. So statement D says shifting f(x) to the left by π/6 gives g(x), which is correct. Wait, but wait, wait. Let me confirm:Original f(x) = √2 sin(2x + π/6)Shift left by π/6: replace x with x + π/6:f(x + π/6) = √2 sin[2(x + π/6) + π/6] = √2 sin(2x + π/3 + π/6) = √2 sin(2x + π/2) = √2 cos(2x) = g(x). So yes, correct. So D is true.Wait, but according to this, shifting left by π/6 gives g(x). So statement D is correct. But wait, hold on, the original question says "Shifting the graph of f(x) to the left by π/6 units will result in the graph of g(x)". So shifting left by π/6, which is indeed replacing x with x + π/6, which gives us g(x). Therefore, D is correct.Wait, but wait. Hold on. If you shift a function to the left by c units, the transformation is h(x) = f(x + c). So here, shifting left by π/6 would result in f(x + π/6). Which we just saw equals g(x). So D is correct.But in the options, A is false, B is true, C is true, D is true. Wait, but the options are A, B, C, D. Let me check again.Wait, no. Wait, when we did the calculation for D, we found that shifting f(x) left by π/6 gives g(x). So D is correct. So then, the correct options are B, C, D. But wait, the problem is likely from a Chinese exam, and such questions often have multiple correct options, but sometimes only one. Wait, but the way the question is phrased, "then ()", with options. Let me check again.Wait, let me confirm each step again to avoid mistakes.Starting with D: shifting f(x) left by π/6. Let's verify with specific x values.Take x = 0. Then f(0) = √2 sin(π/6) = √2*(1/2) = √2/2. Shifting left by π/6, we have f(0 + π/6) = f(π/6) = √2 sin(2*(π/6) + π/6) = √2 sin(π/3 + π/6) = √2 sin(π/2) = √2*1 = √2. Which is equal to g(0) = √2 cos(0) = √2*1 = √2. So that works.Another point: x = π/6. Original f(π/6) = √2 sin(2*(π/6) + π/6) = √2 sin(π/3 + π/6) = √2 sin(π/2) = √2. Shift left by π/6: f(π/6 + π/6) = f(π/3) = √2 sin(2*(π/3) + π/6) = √2 sin(2π/3 + π/6) = √2 sin(5π/6) = √2*(1/2) = √2/2. Which should equal g(π/6) = √2 cos(2*(π/6)) = √2 cos(π/3) = √2*(1/2) = √2/2. Correct. So D seems correct.But wait, the original problem says "shifting the graph of f(x) to the left by π/6 units will result in the graph of g(x)". Which is exactly what we've shown. So D is correct.Wait, but in the options, are B, C, D correct? Let me check again.- A: False- B: True- C: True- D: TrueBut in the problem statement, the options are labeled A, B, C, D. So possible answers could be B, C, D. But in the original problem, maybe only one is correct? Wait, but the problem might be multiple correct. Let me check the problem again.The problem says "then ()", followed by options A to D. In Chinese exams, typically multiple choices can be correct, but the way the options are presented here isn't clear. However, the instruction here says to put the final answer within boxed{}, which usually indicates a single answer. But in the original problem, perhaps multiple options are correct, and we need to box all correct ones. However, in the context of the user's question, since they're asking to put the final answer within boxed{}, and the options are labeled with uppercase letters, perhaps multiple letters need to be boxed. But let me check.Wait, looking back at the original problem:"Given functions... both of which have a minimum positive period of π... then ()A: The maximum value of f(x)+g(x) is 2B: The graph... symmetric about the point...C: Both f(x) and g(x) are monotonically increasing...D: Shifting the graph of f(x) to the left..."In Chinese standardized tests, often each option is a separate possibility, and the correct answer is all that apply, indicated by multiple boxed letters. But since the user's instruction says "put your final answer within boxed{}", and typically in problems here, even with multiple correct answers, they are boxed together, like boxed{BCD}. So based on the analysis, the correct options are B, C, D. But let me confirm once more.Wait, for option C: Both f and g are increasing on [2π/3, 5π/6]. Let's recheck.For f(x) = √2 sin(2x + π/6), derivative f’(x) = 2√2 cos(2x + π/6). On [2π/3, 5π/6], 2x + π/6 ranges from 4π/3 + π/6 = 9π/6 = 3π/2 to 5π/3 + π/6 = 11π/6. As discussed earlier, cosine is positive in this interval (from 3π/2 to 11π/6), so f’(x) is positive. Therefore, f is increasing.For g(x) = √2 cos(2x), derivative g’(x) = -2√2 sin(2x). On [2π/3, 5π/6], 2x ranges from 4π/3 to 5π/3. The sine of angles between 4π/3 and 5π/3 is negative (since 4π/3 is 240°, 5π/3 is 300°, both in third and fourth quadrants where sine is negative). Therefore, sin(2x) is negative, so g’(x) = -2√2*(negative) = positive. So g is increasing. Therefore, C is correct.Therefore, all of B, C, D are correct. So the answer should be B, C, D.But wait, let me check the problem statement again. The original problem says "both of which have a minimum positive period of π". We determined ω = 2. That's correct. Then, given that f(x) + g(x) has a zero at -π/6, which led us to φ = π/6. Then, the options.But let me check for option D again. We found that shifting f(x) left by π/6 gives g(x). But wait, in general, shifting left by c units is a horizontal shift, which for a function h(x), shifted left by c is h(x + c). So in this case, f(x + π/6) = g(x). So for all x, f(x + π/6) = g(x). Therefore, the graph of f shifted left by π/6 is indeed g(x). So D is correct.Therefore, the correct options are B, C, D. However, in the original problem statement, the options are labeled A, B, C, D. The user hasn't specified if multiple answers are possible, but given the context, likely. In Chinese exams, this is common.But looking back at the initial problem translation: the user wrote "then ()", with options A to D. In Chinese exams, multiple can be correct, and you select all that apply. So the answer is B, C, D.But let's cross-verify once more.Wait, regarding statement B: symmetric about the point (-2π/3, 0). We found that h(x) = sqrt(6) sin(2x + π/3). We checked that h(-4π/3 - x) = -h(x), which is the condition for point reflection about (-2π/3, 0). Therefore, B is correct.Therefore, all except A are correct. So the answer is B, C, D.But I need to check if the problem allows multiple answers. Since the user hasn't specified, but given the options are labeled A to D, and in Chinese tests typically multiple can be correct, so the answer is B, C, D. Thus, the boxed answer should be boxed{BCD}.Final Answerboxed{BCD}</think>Given functions ( f(x) = sqrt{2} sin(omega x + varphi) ) and ( g(x) = sqrt{2} cos(omega x) ) with (omega > 0) and (varphi in [0, pi)), both having a minimum positive period of (pi), and a zero point of ( f(x) + g(x) ) at ( -frac{pi}{6} ):1. Determine (omega) and (varphi): - The period of both functions is (pi), so (omega = 2). - Using the zero point ( x = -frac{pi}{6} ), we solve ( sin(varphi - frac{pi}{3}) = -frac{1}{2} ), leading to (varphi = frac{pi}{6}).2. Check the statements: - A: The maximum value of ( f(x) + g(x) ) is calculated as ( sqrt{6} ), not 2. Hence, A is false. - B: The function ( f(x) + g(x) = sqrt{6} sin(2x + frac{pi}{3}) ) is symmetric about ((- frac{2pi}{3}, 0)). Verified by checking ( h(-4pi/3 - x) = -h(x) ). Hence, B is true. - C: Both ( f(x) ) and ( g(x) ) are monotonically increasing on ([ frac{2pi}{3}, frac{5pi}{6} ]). Verified by checking derivatives. Hence, C is true. - D: Shifting ( f(x) ) left by (frac{pi}{6}) results in ( g(x) ). Verified by substitution. Hence, D is true.Final answer: (boxed{BCD})]

🔑:Given the functions f(x)=sqrt{2}sin(omega x+varphi) and g(x)=sqrt{2}cos(omega x), with omega > 0 and varphi in [0,pi), both having a minimum positive period of pi, and a zero point of f(x)+g(x) at x=-frac{pi}{6}, we are to determine the correct statements among the given options.Step 1: Determine omegaGiven that both functions have a minimum positive period of pi, we can deduce the value of omega as follows:[frac{2pi}{omega} = pi implies omega = 2.]Step 2: Express f(x)+g(x)Substituting omega = 2 into f(x) and g(x), we get:[f(x)+g(x) = sqrt{2}sin(2x+varphi) + sqrt{2}cos(2x).]Step 3: Use the zero point to find varphiGiven that f(x)+g(x) has a zero point at x=-frac{pi}{6}, we substitute x into f(x)+g(x):[sqrt{2}sinleft(-frac{pi}{3}+varphiright) + sqrt{2}cosleft(-frac{pi}{3}right) = 0.]This simplifies to:[sin(varphi-frac{pi}{3}) = -frac{1}{2}.]Given varphi in [0,pi), we find:[varphi-frac{pi}{3} = -frac{pi}{6} implies varphi = frac{pi}{6}.]Step 4: Re-express f(x)+g(x) with found varphiWith varphi = frac{pi}{6}, we have:[f(x)+g(x) = sqrt{2}sin(2x+frac{pi}{6}) + sqrt{2}cos(2x) = sqrt{6}left(frac{1}{2}sin(2x) + frac{sqrt{3}}{2}cos(2x)right) = sqrt{6}sin(2x+frac{pi}{3}).]Step 5: Analyze the options- Option A: The maximum value of f(x)+g(x) is sqrt{6}, not 2. Thus, option A is incorrect.- Option B: Given f(-frac{2pi}{3})+g(-frac{2pi}{3})=0, the graph of f(x)+g(x) is symmetric about the point (-frac{2pi}{3},0). Thus, option B is correct.- Option C: For f(x) and g(x) within [frac{2pi}{3},frac{5pi}{6}], both functions are monotonically increasing. Thus, option C is correct.- Option D: Shifting the graph of f(x) to the left by frac{pi}{6} units results in g(x). Thus, option D is correct.Therefore, the correct options are boxed{BCD}.

🧠:[<think>Okay, so I need to find the radius of the inscribed circle in a rhombus ABCD. Let me first recall that in a rhombus, all sides are equal, and the radius of the inscribed circle is equal to the height of the rhombus divided by 2. So, if I can find the height, then dividing by 2 will give me the radius. Alternatively, since the area of the rhombus can also be expressed as base times height or as (product of the diagonals)/2, maybe I can find the area through some other means and then divide by the perimeter to get the radius. Wait, actually, the formula for the radius of an inscribed circle in a rhombus is area divided by perimeter. So, radius r = Area / Perimeter. Since all sides are equal, perimeter is 4 times a side. So, if I can find the area and the side length, that would work. But let's see what information we have here.We have a rhombus ABCD. Point Q divides side BC in the ratio 1:3 from B, so BQ:QC = 1:3. Point E is the midpoint of AB. In triangle CEQ, the median CF is 2√2, and EQ is √2. So, let's try to visualize this.First, let me sketch the rhombus. Let me consider ABCD with AB, BC, CD, DA as sides. Since it's a rhombus, all sides are equal. Let me denote the side length as 's'. Let me place the rhombus in a coordinate system to make things clearer. Let me assume point A is at the origin (0,0), and since it's a rhombus, if I consider AB along the x-axis, then point B would be at (s, 0). Then, point D would be at (0, h), where h is the height of the rhombus. Wait, no, actually, that might not be accurate because the diagonals of a rhombus bisect each other at right angles. Hmm, maybe using coordinates is a good approach here. Let me try to set up coordinates for all the points.Let me place point A at (0,0). Let me assume that the rhombus is such that AB is along the x-axis, so point B is at (a, 0) for some a. Since it's a rhombus, all sides are of length a. Then, point D would be at (0, b), and point C would be at (a, b), but wait, that would make it a rectangle. Hmm, no, in a rhombus, the sides are equal, but the angles can vary. Maybe a better coordinate system is to have point A at (0,0), point B at (s,0), point C at (s + t, u), and point D at (t, u), where s, t, u are parameters. But this might complicate things. Alternatively, perhaps using vectors or coordinate geometry with the center at the origin.Alternatively, since the problem involves medians and midpoints, coordinate geometry might be the way to go. Let me proceed with coordinates.Let me set up coordinates such that the center of the rhombus is at the origin (0,0). Then, the vertices can be represented as follows: if the diagonals are of lengths 2p and 2q, then the vertices are at (p,0), (-p,0), (0,q), (0,-q). But in this case, the sides are not along the axes, but the diagonals are. However, since the problem refers to sides AB, BC, etc., maybe it's better to have sides along the axes. Wait, no, in a rhombus, the diagonals bisect each other at right angles. So, if I set the rhombus such that its diagonals are along the coordinate axes, then the vertices would be at (p,0), (0,q), (-p,0), (0,-q). But then the sides are from (p,0) to (0,q), etc. The length of each side would be sqrt(p^2 + q^2), so all sides are equal. Then, the area is 2p*2q / 2 = 2pq. The radius of the inscribed circle would be area / perimeter = (2pq) / (4*sqrt(p^2 + q^2)) ) = pq / (2*sqrt(p^2 + q^2)). Hmm, but maybe this is complicating things. Alternatively, perhaps setting up the rhombus with sides along the coordinate system.Wait, perhaps another approach. Let me let AB be along the x-axis, with point A at (0,0), point B at (s,0). Then, since it's a rhombus, point D must be at some point (d, e), such that the length AD = AB = s. So, sqrt(d^2 + e^2) = s. Then, point C would be at point B + vector AD, which is (s + d, e). Then, the coordinates would be:A: (0,0)B: (s,0)D: (d,e)C: (s + d, e)This way, all sides are length s. So, AB is from (0,0) to (s,0), BC is from (s,0) to (s + d, e), CD is from (s + d, e) to (d, e), and DA is from (d, e) back to (0,0). All sides have length sqrt(d^2 + e^2) = s.Now, point Q divides BC in the ratio 1:3 from B. So, from point B to point C, the vector is (d, e). So, point Q is located 1/4 of the way from B to C. Therefore, the coordinates of Q are:Q_x = s + (1/4)*dQ_y = 0 + (1/4)*e = e/4Similarly, point E is the midpoint of AB, so E is at (s/2, 0).Now, triangle CEQ. Let's find the coordinates of C, E, Q.C is at (s + d, e)E is at (s/2, 0)Q is at (s + d/4, e/4)So, we need to consider triangle CEQ with these three points.The median CF of triangle CEQ is the median from point C to side EQ. The median CF connects point C to the midpoint F of EQ.First, let's find the coordinates of F, the midpoint of EQ.Coordinates of E: (s/2, 0)Coordinates of Q: (s + d/4, e/4)So, midpoint F has coordinates:F_x = (s/2 + s + d/4)/2 = (3s/2 + d/4)/2 = (3s/2)/2 + (d/4)/2 = 3s/4 + d/8F_y = (0 + e/4)/2 = e/8Therefore, F is at (3s/4 + d/8, e/8)Now, the median CF connects point C (s + d, e) to point F (3s/4 + d/8, e/8). The length of CF is given as 2√2.So, let's compute the distance between C and F.Difference in x-coordinates: (3s/4 + d/8) - (s + d) = 3s/4 - s + d/8 - d = (-s/4) - (7d/8)Difference in y-coordinates: e/8 - e = -7e/8Therefore, distance CF is sqrt[ (-s/4 - 7d/8)^2 + (-7e/8)^2 ] = 2√2Similarly, EQ is given as √2. Let's compute the length of EQ.Coordinates of E: (s/2, 0)Coordinates of Q: (s + d/4, e/4)Difference in x: s + d/4 - s/2 = s/2 + d/4Difference in y: e/4 - 0 = e/4Therefore, EQ = sqrt[ (s/2 + d/4)^2 + (e/4)^2 ] = √2So, we have two equations:1) sqrt[ (-s/4 - 7d/8)^2 + (-7e/8)^2 ] = 2√22) sqrt[ (s/2 + d/4)^2 + (e/4)^2 ] = √2Additionally, since ABCD is a rhombus, all sides are equal, so the length from A to B is s, and the length from A to D is sqrt(d^2 + e^2) = s. Therefore, we have:3) d^2 + e^2 = s^2So, we have three equations with three variables: s, d, e.Let me square both sides of equations 1 and 2 to eliminate the square roots.Starting with equation 2:[ (s/2 + d/4)^2 + (e/4)^2 ] = (√2)^2 = 2Similarly, equation 1:[ (-s/4 - 7d/8)^2 + (-7e/8)^2 ] = (2√2)^2 = 8So, let's write them out:Equation 2:( (s/2 + d/4)^2 + (e/4)^2 ) = 2Equation 1:( (-s/4 - 7d/8)^2 + ( -7e/8 )^2 ) = 8Let me expand these equations.Starting with equation 2:First term: (s/2 + d/4)^2 = (s/2)^2 + 2*(s/2)*(d/4) + (d/4)^2 = s²/4 + (s d)/4 + d²/16Second term: (e/4)^2 = e²/16So, equation 2 becomes:s²/4 + (s d)/4 + d²/16 + e²/16 = 2Multiply all terms by 16 to eliminate denominators:4s² + 4s d + d² + e² = 32Equation 3 is d² + e² = s². So, substitute d² + e² with s²:4s² + 4s d + s² = 32Simplify:5s² + 4s d = 32Let's note that as equation 2a: 5s² + 4s d = 32Now, equation 1:First term: (-s/4 -7d/8)^2 = (s/4 +7d/8)^2 = (s/4)^2 + 2*(s/4)*(7d/8) + (7d/8)^2 = s²/16 + (14 s d)/32 + 49 d² /64 = s²/16 + (7 s d)/16 + 49 d² /64Second term: (-7e/8)^2 = 49 e² /64So, equation 1 becomes:s²/16 + (7 s d)/16 + 49 d² /64 + 49 e² /64 = 8Combine terms:Multiply all terms by 64 to eliminate denominators:4 s² + 28 s d + 49 d² + 49 e² = 512From equation 3, we know that e² = s² - d². Substitute e²:4 s² + 28 s d + 49 d² + 49 (s² - d²) = 512Simplify:4 s² + 28 s d + 49 d² + 49 s² - 49 d² = 512Combine like terms:(4s² + 49s²) + 28 s d + (49 d² -49 d²) = 51253 s² + 28 s d = 512So, equation 1a: 53 s² + 28 s d = 512Now, we have two equations:2a: 5s² + 4s d = 321a: 53s² + 28s d = 512We can solve this system for s² and s d.Let me denote x = s² and y = s d.Then, equation 2a: 5x + 4y = 32Equation 1a: 53x + 28y = 512Let me solve these equations.Multiply equation 2a by 7: 35x + 28y = 224Subtract equation 1a:(35x + 28y) - (53x + 28y) = 224 - 512-18x = -288Therefore, x = (-288)/(-18) = 16So, x = s² = 16 => s = 4 (since length is positive)Now, substitute x =16 into equation 2a: 5*16 +4y = 32 => 80 +4y=32 => 4y= -48 => y= -12But y = s d = 4*d = -12 => d= -12/4= -3So, d= -3, s=4.Now, from equation 3: d² + e² = s² => (-3)^2 + e² =16 => 9 + e²=16 => e²=7 => e=±√7. Since we're dealing with coordinates, the sign depends on the orientation. Since we can take e positive (as it's a height), let's take e=√7.So, now we have s=4, d=-3, e=√7.Now, let's check these values in equation 1a:53x +28y =53*16 +28*(-12)=848 -336=512. Correct.So, the side length s=4, and d=-3, e=√7.Now, we need to find the radius of the inscribed circle in the rhombus.As mentioned earlier, the radius is equal to the area divided by the perimeter.Perimeter of the rhombus is 4*s=4*4=16.Area of the rhombus can be calculated in several ways. Since we know the diagonals, but we can also compute it as base*height.But let's compute the area.Alternatively, in the coordinate system, the area can be calculated using vectors or the shoelace formula. Let's use the shoelace formula for the coordinates.But wait, in the coordinates set up earlier, points are:A: (0,0)B: (4,0)D: (-3,√7)C: (4 + (-3), √7) = (1,√7)Wait, is that correct? Wait, point C is (s + d, e). Since s=4, d=-3, so 4 + (-3)=1, and e=√7. So, C is (1,√7). Let me confirm the coordinates:A: (0,0)B: (4,0)C: (1,√7)D: (-3,√7)Wait, but then AD is from (0,0) to (-3,√7), which should have length sqrt( (-3)^2 + (√7)^2 )=sqrt(9 +7)=sqrt(16)=4, which matches s=4. Similarly, AB is from (0,0) to (4,0), length 4. BC is from (4,0) to (1,√7), which is sqrt( (-3)^2 + (√7)^2 )=sqrt(9 +7)=sqrt(16)=4. Correct.So, the rhombus is set with these coordinates.Now, area of the rhombus can be calculated as the magnitude of the cross product of vectors AB and AD.Vector AB is (4,0), vector AD is (-3,√7). The cross product is (4)(√7) - (0)(-3) =4√7. So, the area is |4√7|=4√7.Alternatively, using the shoelace formula:Coordinates of the rhombus in order A(0,0), B(4,0), C(1,√7), D(-3,√7), back to A(0,0).Compute shoelace sum:Sum1 = (0*0) + (4*√7) + (1*√7) + (-3*0) = 0 +4√7 +√7 +0=5√7Sum2 = (0*4) + (0*1) + (√7*(-3)) + (√7*0)=0 +0 -3√7 +0= -3√7Area= |Sum1 - Sum2| /2 = |5√7 - (-3√7)| /2 = |8√7| /2 =4√7. Correct.So, area is 4√7, perimeter is 16. Therefore, radius r=Area / Perimeter=4√7 /16=√7 /4.Wait, but this is conflicting with my initial thought that the radius is height divided by 2. Let's check.In a rhombus, the radius of the inscribed circle is equal to the height (distance between two opposite sides) divided by 2? Wait, actually, no. Wait, in a rhombus, the area is base times height. Since all sides are equal, the area is s * h, where h is the height. Then, the radius is area / perimeter = (s h) / (4 s) )= h /4. Wait, so radius is h /4. But according to the area formula, if area is s h, and perimeter is 4 s, then radius r = (s h)/(4 s)=h /4. So, the radius is h/4, not h/2. So, that contradicts my initial thought, but according to the formula, yes.Alternatively, maybe I was confusing with another polygon. Anyway, according to the calculation, the radius is √7 /4. But let me confirm with the given problem.Wait, but according to our coordinates, the height can be calculated. Let's see. The height is the distance between sides AB and CD. Since AB is along the x-axis from (0,0) to (4,0), and CD is from (1,√7) to (-3,√7). The y-coordinate of CD is √7, so the height is the vertical distance between AB (y=0) and CD (y=√7). Wait, but that is √7. Therefore, the height h=√7. Then, the radius should be h /4=√7 /4. Which matches the previous result.So, the radius is √7 /4. Therefore, the answer is √7 /4, which is equal to (1/4)√7. To write it in boxed form, boxed{dfrac{sqrt{7}}{4}}.But let me just verify all steps once more to ensure there are no errors.First, setting up coordinates correctly? Yes, with A at (0,0), B at (4,0), D at (-3,√7), C at (1,√7). Then, equations for Q and E. Point Q is 1/4 from B to C, so coordinates (4 + (1/4)(-3), 0 + (1/4)(√7))= (4 - 3/4, √7/4)= (13/4, √7/4). Wait, hold on. Wait, earlier I had point Q as (s + d/4, e/4). With s=4, d=-3, e=√7. So, Q is (4 + (-3)/4, √7 /4)= (4 - 3/4, √7 /4)= (13/4, √7 /4). Similarly, point E is midpoint of AB: (2,0).Then, triangle CEQ. C is (1,√7), E is (2,0), Q is (13/4, √7 /4). Let's check the coordinates again.Wait, C is (1,√7). E is (2,0). Q is (13/4, √7 /4).Then, midpoint F of EQ: E is (2,0), Q is (13/4, √7 /4). Midpoint F is ((2 +13/4)/2, (0 + √7 /4)/2)= ((8/4 +13/4)/2, (√7 /4)/2)= (21/4 /2, √7 /8)= (21/8, √7 /8). So, F is (21/8, √7 /8). Then, the median CF is from C(1,√7) to F(21/8, √7 /8). Let's compute the distance between these two points.Difference in x: 21/8 -1=21/8 -8/8=13/8Difference in y: √7 /8 -√7= √7 /8 -8√7 /8= (-7√7)/8So, distance CF is sqrt( (13/8)^2 + (-7√7 /8)^2 )Compute:(13/8)^2=169/64(-7√7 /8)^2=49*7 /64=343/64Sum:169/64 +343/64=512/64=8sqrt(8)=2√2. Correct, which matches the given CF=2√2.Similarly, compute EQ. E(2,0) to Q(13/4, √7 /4). Difference in x:13/4 -2=13/4 -8/4=5/4. Difference in y:√7 /4 -0=√7 /4. So, EQ= sqrt( (5/4)^2 + (√7 /4)^2 )=sqrt(25/16 +7/16)=sqrt(32/16)=sqrt(2)=√2. Correct. So, all given lengths are satisfied.Therefore, the calculations are correct. The side length s=4, height h=√7, so the radius r= h /4=√7 /4.Therefore, the answer is boxed{dfrac{sqrt{7}}{4}}.Final Answerboxed{dfrac{sqrt{7}}{4}}</think>Given a rhombus (ABCD) with point (Q) dividing side (BC) in the ratio (1:3) from (B), and point (E) as the midpoint of (AB). The median (CF) of triangle (CEQ) is (2sqrt{2}), and (EQ = sqrt{2}). We need to find the radius of the inscribed circle in rhombus (ABCD).1. Coordinate Setup: - Place (A) at ((0,0)), (B) at ((s,0)), (D) at ((d,e)), and (C) at ((s+d, e)). - All sides are equal, so (s = sqrt{d^2 + e^2}).2. Coordinates of Points: - (E) (midpoint of (AB)) is ((s/2, 0)). - (Q) (dividing (BC) in ratio (1:3)) is ((s + frac{d}{4}, frac{e}{4})).3. Equations from Given Lengths: - Midpoint (F) of (EQ) is (left(frac{3s}{4} + frac{d}{8}, frac{e}{8}right)). - Distance (CF = 2sqrt{2}) and (EQ = sqrt{2}).4. Solving Equations: - Using the coordinates and given lengths, derive equations: [ sqrt{left(frac{s}{2} + frac{d}{4}right)^2 + left(frac{e}{4}right)^2} = sqrt{2} ] [ sqrt{left(-frac{s}{4} - frac{7d}{8}right)^2 + left(-frac{7e}{8}right)^2} = 2sqrt{2} ] - Solve these along with (d^2 + e^2 = s^2).5. Solution: - Find (s = 4), (d = -3), (e = sqrt{7}). - Area of rhombus (= 4sqrt{7}). - Perimeter of rhombus (= 16).6. Radius Calculation: - Radius (r = frac{text{Area}}{text{Perimeter}} = frac{4sqrt{7}}{16} = frac{sqrt{7}}{4}).Thus, the radius of the inscribed circle is (boxed{dfrac{sqrt{7}}{4}}).,<think>Okay, let's try to tackle this problem step by step. It's about a rhombus with some points and lengths given, and we need to find the radius of the inscribed circle. Hmm, first, let me recall that in a rhombus, all sides are equal, the diagonals bisect each other at right angles, and the radius of the inscribed circle is equal to half the height of the rhombus. So, if I can find the area of the rhombus and divide it by its perimeter, that will give me the radius. The formula is r = Area / Perimeter. Since all sides are equal, perimeter is 4 times a side. But maybe there's another way using the given information about the triangle and median. Let me digest the problem again.We have rhombus ABCD. Point Q divides BC in the ratio 1:3 from B, so BQ:QC = 1:3. Point E is the midpoint of AB. Then, in triangle CEQ, the median CF is 2√2, and EQ is √2. Need to find the inradius.First, maybe it's helpful to assign coordinates to the rhombus. Let me try to set up a coordinate system. Let's place the rhombus such that its center is at the origin, and the diagonals are along the axes. Wait, but maybe that complicates things because we have points E and Q. Alternatively, maybe place vertex A at the origin, side AB along the x-axis. Let's see.Let me try this approach. Let’s denote the rhombus ABCD with AB, BC, CD, DA. All sides are equal. Let's let AB be along the x-axis. Let’s assign coordinates:Let’s let point A be at (0, 0). Since it's a rhombus, all sides are equal. Let's denote the side length as 's'. Let’s denote angle at A as θ, so the coordinates can be determined using θ. But maybe using coordinates with vectors or something. Wait, maybe let's assume coordinates more explicitly.Let’s set point A at (0,0). Since E is the midpoint of AB, then if AB is from (0,0) to (s,0), then E is at (s/2, 0). Then point B is (s, 0). Now, since it's a rhombus, point D is at (0, h), where h is the height? Wait, no. In a rhombus, the height is the distance between two opposite sides. If we have sides of length s, the height h is s*sinθ, where θ is any interior angle. But coordinates might not be straightforward unless we use vectors.Alternatively, let's define the rhombus with coordinates:Let’s let point A be at (0,0). Then, since it's a rhombus, suppose point B is at (a,0), point D is at (0,b), then point C would be at (a,b). But in a rhombus, all sides are equal, so the distance from A to B is a, and from B to C is sqrt((a - a)^2 + (b - 0)^2) = b. So for all sides to be equal, we need a = b. Wait, no. Wait, if points are A(0,0), B(a,0), C(a + c, d), D(c, d). Then sides AB, BC, CD, DA must all be equal. Maybe this is getting complicated. Let's see.Alternatively, consider the rhombus with vertices at A(0,0), B(b,0), C(b + c, d), D(c, d). Then, the sides AB, BC, CD, DA must have the same length. So AB length is sqrt((b)^2 + 0) = b. BC is sqrt((c)^2 + (d)^2). CD is sqrt(( -b)^2 + 0) = b. DA is sqrt((c - b - c)^2 + (d - 0)^2) = sqrt((-b)^2 + d^2). So to have all sides equal, we need sqrt(c² + d²) = b. Hmm, so if we set c and d such that c² + d² = b², then all sides are equal. Maybe this is a way to parametrize the rhombus.But maybe this is getting too involved. Alternatively, let's use vectors. Let me think.Let’s take point A at the origin. Let vector AB be along the x-axis, so vector AB = (s, 0), where s is the side length. Then, vector AD would be (s*cosθ, s*sinθ), where θ is the angle at vertex A. Then coordinates:A: (0, 0)B: (s, 0)D: (s*cosθ, s*sinθ)C: B + AD = (s + s*cosθ, 0 + s*sinθ) = (s(1 + cosθ), s sinθ)Wait, but in a rhombus, the diagonals bisect each other. Alternatively, maybe better to have coordinates based on half diagonals. Let me recall that in a rhombus, area is (d1 * d2)/2, where d1 and d2 are the diagonals. Also, the inradius is area divided by perimeter, so r = (d1*d2)/2 / (4s) = (d1*d2)/(8s). But maybe we can relate diagonals to the sides. Since diagonals satisfy (d1/2)^2 + (d2/2)^2 = s^2. So, maybe if I can find the diagonals, then compute area and perimeter.But perhaps I should proceed with coordinates. Let me try again.Set A at (0,0). Let’s let the rhombus have sides of length 's'. Let’s take AB along the x-axis, so point B is (s,0). Then, point D is at (p, q), so that AD is (p, q), and since all sides are equal, the length AD is s: sqrt(p² + q²) = s. Then point C is B + D - A = (s + p, q). So coordinates:A: (0,0)B: (s, 0)D: (p, q)C: (s + p, q)Then, the midpoint of AB, point E, is ((0 + s)/2, (0 + 0)/2) = (s/2, 0).Point Q divides BC in the ratio 1:3 from B. Let's find coordinates of Q.First, BC is from B(s, 0) to C(s + p, q). The ratio BQ:QC = 1:3, so Q divides BC into four parts, with BQ being 1 part and QC being 3 parts. So using the section formula, the coordinates of Q are:Q_x = (3*s + 1*(s + p)) / (1 + 3) = (3s + s + p)/4 = (4s + p)/4 = s + p/4Q_y = (3*0 + 1*q)/4 = q/4So Q is (s + p/4, q/4)Now, triangle CEQ. Let's find coordinates of C, E, Q.C is (s + p, q)E is (s/2, 0)Q is (s + p/4, q/4)So, points of triangle CEQ:C: (s + p, q)E: (s/2, 0)Q: (s + p/4, q/4)We need to find the median CF of triangle CEQ, where F is the midpoint of EQ. Let's find coordinates of F first.Midpoint F of EQ:E is (s/2, 0)Q is (s + p/4, q/4)So F_x = (s/2 + s + p/4)/2 = ( (3s/2) + p/4 ) / 2 = (3s/2 + p/4)/2 = (3s/4 + p/8)F_y = (0 + q/4)/2 = q/8Therefore, F is (3s/4 + p/8, q/8)Now, the median CF connects point C(s + p, q) to F(3s/4 + p/8, q/8). Let's compute the length of CF, which is given as 2√2.Compute the distance between C and F:Δx = (3s/4 + p/8) - (s + p) = (3s/4 - s) + (p/8 - p) = (-s/4) + (-7p/8) = -s/4 -7p/8Δy = q/8 - q = -7q/8Therefore, the distance CF is sqrt[ (-s/4 -7p/8)^2 + (-7q/8)^2 ] = 2√2.Similarly, EQ is given as √2. Let's compute EQ.Points E(s/2, 0) and Q(s + p/4, q/4). Distance EQ:Δx = s + p/4 - s/2 = s/2 + p/4Δy = q/4 - 0 = q/4So EQ = sqrt[ (s/2 + p/4)^2 + (q/4)^2 ] = √2So now, we have two equations:1) sqrt[ (-s/4 -7p/8)^2 + (-7q/8)^2 ] = 2√22) sqrt[ (s/2 + p/4)^2 + (q/4)^2 ] = √2Also, since ABCD is a rhombus, we have that the length of AD is s, so sqrt(p^2 + q^2) = s. So equation 3:3) p² + q² = s²So, we have three equations with variables p, q, s.Let me square both sides of equations 1 and 2 to eliminate the square roots.Equation 1 squared:[ (-s/4 -7p/8 )² + (-7q/8 )² ] = (2√2)² = 8Compute the terms inside:First term: (-s/4 -7p/8 )² = (s/4 +7p/8 )² = ( (2s +7p)/8 )² = (2s +7p)² / 64Second term: (-7q/8 )² = 49 q² / 64Therefore, equation 1 becomes:( (2s +7p)^2 + 49 q² ) / 64 = 8Multiply both sides by 64:(2s +7p)^2 + 49 q² = 512Similarly, equation 2 squared:[ (s/2 + p/4 )² + (q/4 )² ] = 2Compute terms:First term: (s/2 + p/4 )² = ( (2s + p)/4 )² = (2s + p)² / 16Second term: (q/4 )² = q² / 16Thus, equation 2 becomes:( (2s + p)^2 + q² ) / 16 = 2Multiply both sides by 16:(2s + p)^2 + q² = 32So now, we have:Equation 1a: (2s +7p)^2 +49 q² = 512Equation 2a: (2s + p)^2 + q² = 32Equation 3: p² + q² = s²Hmm, three equations with three variables s, p, q.Let me denote equations 2a and 3. Let me try to express q² from equation 3: q² = s² - p². Then substitute into equation 2a.Equation 2a becomes:(2s + p)^2 + (s² - p²) = 32Expand (2s + p)^2: 4s² +4sp + p²So:4s² +4sp + p² + s² - p² = 32Simplify:4s² +4sp + s² = 32 => 5s² +4sp = 32So equation 4: 5s² +4sp =32Similarly, let's handle equation 1a. Substitute q² = s² - p² into equation 1a.Equation 1a: (2s +7p)^2 +49(s² - p²) =512Compute (2s +7p)^2: 4s² +28sp +49p²So:4s² +28sp +49p² +49s² -49p² =512Simplify:4s² +49s² +28sp +49p² -49p² =512Which is:53s² +28sp =512So equation 5:53s² +28sp=512Now, from equation 4:5s² +4sp=32, let's solve for sp.Equation 4: 5s² +4sp=32 => 4sp=32 -5s² => sp= (32 -5s²)/4Now, substitute sp into equation 5.Equation 5:53s² +28sp=512Substitute sp=(32 -5s²)/4:53s² +28*(32 -5s²)/4 =512Simplify 28/4=7:53s² +7*(32 -5s²)=512Compute 7*32=224, 7*(-5s²)= -35s²So:53s² -35s² +224=512 =>18s² +224=512Subtract 224:18s²=512-224=288 =>s²=288/18=16 =>s=4 (since s is positive)So s=4.Then, from equation 4:5*(16) +4sp=32 =>80 +4sp=32 =>4sp=32-80= -48 =>sp= -12So sp= -12. Since s=4, p= -12/4= -3.Thus, p= -3.Now, from equation 3: p² + q² =s² => (-3)^2 + q²=16 =>9 + q²=16 =>q²=7 =>q=√7 or -√7, but since q is a coordinate, we can take positive q=√7.Therefore, we have s=4, p=-3, q=√7.So now, the rhombus has vertices:A(0,0), B(4,0), D(-3,√7), C(4 + (-3),√7)= (1,√7)Wait, wait. Wait, from earlier, point C was (s + p, q). Since s=4, p=-3, so C(4 + (-3), √7)= (1, √7). Let me confirm coordinates:A: (0,0)B: (4,0)D: (-3,√7)C: B + D - A = (4 + (-3),0 + √7)= (1, √7)Yes. So coordinates are:A(0,0); B(4,0); C(1,√7); D(-3,√7)Now, need to find the radius of the inscribed circle in the rhombus. As mentioned earlier, the radius is equal to the area divided by the perimeter.First, compute the perimeter. Since all sides are length s=4, perimeter is 4*4=16.Next, compute the area. In a rhombus, area can be computed as base * height, or (d1*d2)/2, where d1 and d2 are the diagonals.Alternatively, since we have coordinates, we can use the shoelace formula.Let me compute the area using coordinates.Coordinates in order: A(0,0); B(4,0); C(1,√7); D(-3,√7); back to A(0,0).Shoelace formula:Area = 1/2 |sum over edges (x_i y_{i+1} - x_{i+1} y_i)|Compute term by term:A to B: x_i=0, y_i=0; x_{i+1}=4, y_{i+1}=0: contribution 0*0 -4*0=0B to C: x_i=4, y_i=0; x_{i+1}=1, y_{i+1}=√7: contribution 4*√7 -1*0=4√7C to D: x_i=1, y_i=√7; x_{i+1}=-3, y_{i+1}=√7: contribution 1*√7 - (-3)*√7=1√7 +3√7=4√7D to A: x_i=-3, y_i=√7; x_{i+1}=0, y_{i+1}=0: contribution -3*0 -0*√7=0Total sum: 0 +4√7 +4√7 +0=8√7Area =1/2 * |8√7|=4√7Alternatively, since we know the side length is 4, and the height can be calculated as Area / base = (4√7)/4=√7. Therefore, the inradius is equal to the height divided by 2? Wait, wait. Wait, no. Wait, inradius is Area / Perimeter. So inradius r = Area / Perimeter =4√7 /16=√7 /4. Wait, that seems small. Wait, but let's check.Alternatively, let's confirm with another method. Let's compute the height of the rhombus. Since the area is 4√7 and the base is 4, the height h = Area / base =4√7 /4=√7. Therefore, the inradius is equal to the height divided by 2? Wait, no. Wait, in a rhombus, the inradius is actually equal to the height. Wait, no, that doesn't make sense. Wait, inradius is the radius of the circle that fits inside the rhombus, tangent to all sides. For a rhombus, which is a tangential quadrilateral (has an incircle), the inradius is equal to the height of the rhombus divided by 2? Wait, no. Wait, actually, inradius is Area / Perimeter. Since perimeter is 4*4=16, Area is 4√7, so r=4√7 /16=√7 /4.But √7 /4 is approximately 0.66, which seems small. Let me verify.Alternatively, let's compute the inradius another way. In a rhombus, since all sides are equal and it's tangential, the inradius can be calculated as r = (d1*d2)/(2*perimeter), but wait, the formula for area is (d1*d2)/2, so r = Area / Perimeter = (d1*d2)/2 / (4s) = (d1*d2)/(8s). Alternatively, if we compute the height h, since area is base*s, area = s*h => h = Area / s = (4√7)/4=√7. Then, since in a rhombus, the inradius is the height divided by 2? Wait, but no, actually, no. Wait, the inradius is the distance from the center to a side, which is the same as the height divided by 2. Wait, maybe.Wait, let me think. In a rhombus, the inradius is the radius of the circle that fits inside, touching all four sides. The distance from the center of the rhombus to any side is equal to the inradius. Since the rhombus can be divided into four congruent right triangles by its diagonals, the inradius would be the height of one of these triangles from the right angle to the hypotenuse. Alternatively, maybe it's the same as the height of the rhombus divided by 2, since the height is the distance between two opposite sides, and the inradius is the distance from the center to a side, which is half the height. So, if the height h is √7, then the inradius is h/2=√7/2. Wait, but according to the area formula, r = Area / Perimeter =4√7 /16=√7/4. Which contradicts. So which is correct?Wait, let's recall. For a tangential quadrilateral (a quadrilateral that has an incircle), the inradius is given by r = Area / Perimeter. So in this case, Area is 4√7, perimeter is 16, so r=4√7 /16=√7 /4≈0.66. However, if the height is √7≈2.645, then half the height is≈1.322. So clearly, different. Therefore, there must be a mistake in my reasoning.Wait, perhaps I miscalculated the area. Let me verify the area using another method. Since we have coordinates, maybe compute the vectors and use cross product.Vectors AB=(4,0), AD=(-3,√7). The area of the rhombus is the magnitude of the cross product of AB and AD.Cross product in 2D is |Ax*By - Ay*Bx| where A and B are vectors. So AB=(4,0), AD=(-3,√7). The cross product is 4*√7 -0*(-3)=4√7. So area is |4√7|=4√7. So that's correct. Therefore, Area=4√7, perimeter=16, so inradius is 4√7 /16=√7 /4≈0.66. That seems correct.Alternatively, perhaps the height is calculated as Area / base. So base is AB=4, height h=4√7 /4=√7≈2.645. So the height is the distance between sides AB and CD. Then, the inradius is the distance from the center to any side. The center of the rhombus is at the intersection of the diagonals. Let's find the coordinates of the center.The diagonals in the rhombus are AC and BD. Coordinates of A(0,0) and C(1,√7), so midpoint is (0.5, √7/2). Similarly, coordinates of B(4,0) and D(-3,√7), midpoint is (0.5, √7/2). So the center is at (0.5, √7/2). Now, the distance from the center to a side, say side AB, which is along the x-axis from (0,0) to (4,0). The distance from the center (0.5, √7/2) to side AB (which is the line y=0) is just the y-coordinate, which is √7/2≈1.322. But according to r=√7 /4≈0.66, which is half of that. Contradiction. So why is there a discrepancy?Wait, no. Wait, inradius is the distance from the center to the side. Wait, if the center is at (0.5, √7/2), then the distance to side AB (y=0) is √7/2, and the distance to side CD (which is the line passing through points C(1,√7) and D(-3,√7)) is also √7/2. Similarly, distance to other sides. Wait, but then inradius should be √7/2≈1.322. But according to Area / Perimeter, it's √7 /4. So which is correct?Wait, I must have messed up. Let me check again.Area =4√7, perimeter=16. So inradius r= Area / Perimeter=4√7 /16=√7 /4≈0.66. But according to the coordinate calculation, the distance from center to side is √7 /2≈1.322. Therefore, inconsistency. That suggests a mistake in my reasoning.Wait, maybe the formula is different. Wait, inradius is indeed Area / Perimeter for a tangential quadrilateral. But in this case, the rhombus is a tangential quadrilateral, so that formula should hold. Then, if according to coordinates, the distance from center to side is √7 /2, then Area / Perimeter should equal √7 /2. But 4√7 /16=√7 /4. Therefore, discrepancy. So where is the mistake?Wait, maybe my coordinate setup is incorrect. Let me check again.Wait, in my coordinate system, the rhombus is A(0,0); B(4,0); C(1,√7); D(-3,√7). Let me confirm the lengths of the sides.AB: from (0,0) to (4,0): length 4, correct.BC: from (4,0) to (1,√7): Δx= -3, Δy=√7, length sqrt(9 +7)=sqrt(16)=4, correct.CD: from (1,√7) to (-3,√7): Δx= -4, Δy=0, length 4, correct.DA: from (-3,√7) to (0,0): Δx=3, Δy= -√7, length sqrt(9 +7)=sqrt(16)=4, correct. So all sides are 4, correct.Now, the area via shoelace formula was 4√7, correct. So Area =4√7, perimeter=16, so inradius r=4√7 /16=√7 /4.But according to coordinates, the distance from center (0.5, √7/2) to side AB is √7 /2, and to side BC is computed as follows. Let's compute the distance from center to side BC to confirm.Equation of side BC: points B(4,0) and C(1,√7). The line BC can be parametrized. Let's find its equation.Slope of BC: (√7 -0)/(1 -4)=√7 / (-3)= -√7 /3Equation: y -0 = -√7 /3 (x -4)So y= -√7 /3 x +4√7 /3Distance from center (0.5, √7/2) to this line:The formula for distance from point (x0,y0) to ax + by +c=0 is |ax0 + by0 +c| / sqrt(a² +b²)First, rewrite equation in ax + by +c=0:y + (√7 /3)x -4√7 /3 =0 => (√7 /3)x + y -4√7 /3=0Thus, a=√7 /3, b=1, c= -4√7 /3Distance:| (√7 /3)(0.5) +1*(√7 /2) -4√7 /3 | / sqrt( (√7 /3)^2 +1^2 )Compute numerator:(√7 /6) + (√7 /2) - (4√7 /3 )Convert to sixths:(√7 /6) + (3√7 /6) - (8√7 /6 )= (1 +3 -8)√7 /6= (-4√7)/6= -2√7/3Absolute value: 2√7 /3Denominator:sqrt( (7/9) +1 )= sqrt(16/9 )=4/3Therefore, distance is (2√7 /3 ) / (4/3 )= (2√7)/4=√7 /2≈1.322But according to the formula r=Area /Perimeter=√7 /4≈0.6614. So discrepancy. So which one is correct?Wait, inradius is supposed to be the distance from the center to any side. So according to coordinates, it is √7 /2. But according to formula, it's √7 /4. Therefore, contradiction. Therefore, I must have messed up something.Wait, no. Wait, inradius in a tangential quadrilateral is indeed Area / Perimeter. So if that gives √7 /4, but according to coordinates, distance is √7 /2, which one is correct? Let me check with an example.Take a rhombus with side length 4, area 4√7. Then, the inradius is Area / Perimeter=4√7 /16=√7 /4. But if the distance from center to side is √7 /2, then that would suggest that the inradius is √7 /2. Contradiction.Wait, but maybe my area is wrong. Wait, area via shoelace was 4√7, but is that correct?Wait, let's recalculate the area via shoelace:Coordinates in order: A(0,0); B(4,0); C(1,√7); D(-3,√7); A(0,0)Compute sum of x_i y_{i+1} -x_{i+1} y_i:A to B: 0*0 -4*0=0B to C:4*√7 -1*0=4√7C to D:1*√7 - (-3)*√7=1√7 +3√7=4√7D to A: -3*0 -0*√7=0Sum:0 +4√7 +4√7 +0=8√7Area=1/2 |8√7|=4√7. Correct.Alternatively, vectors AB and AD: AB=(4,0), AD=(-3,√7). The area is |AB × AD|=|4*√7 -0*(-3)|=4√7. Correct.So area is indeed 4√7. Therefore, according to formula, inradius is 4√7 /16=√7 /4.But according to coordinates, the distance from the center to the side is √7 /2. Therefore, there's a conflict. Where is the mistake?Wait, maybe my assumption that the center is at (0.5, √7/2) is incorrect? Let me verify.In a rhombus, the diagonals bisect each other. The diagonals are AC and BD.Coordinates of A(0,0); C(1,√7). Midpoint of AC is ((0 +1)/2, (0 +√7)/2)=(0.5, √7/2)Coordinates of B(4,0); D(-3,√7). Midpoint of BD is ((4 + (-3))/2, (0 +√7)/2)=(0.5, √7/2). So yes, the center is at (0.5, √7/2). So that's correct.Then, distance from center to side AB is √7 /2. So why is the formula giving √7 /4?Wait, maybe I misunderstood the formula. Inradius is the radius of the incircle, which touches all four sides. Therefore, the distance from the center to each side is equal to the inradius. But according to coordinates, this distance is √7 /2. But formula says √7 /4. Therefore, one of them is wrong.Wait, maybe I miscalculated the shoelace formula or area. Wait, but area via vectors is 4√7. So with perimeter 16, inradius should be 4√7 /16=√7 /4. But how can the distance from center to side be different?Wait, no, wait. Wait, inradius is Area / Perimeter, which is a general formula for any tangential quadrilateral. Therefore, if the inradius is r=Area / Perimeter, then regardless of the distances, this holds. Therefore, if inradius is √7 /4, but according to coordinates, the distance from center to sides is √7 /2, then there's a mistake.Wait, but in a rhombus, the inradius is equal to the height divided by 2. Wait, let's think. The height is the distance between two opposite sides. In this case, the distance between AB and CD. Since AB is at y=0 and CD is at y=√7 (because points C and D are at y=√7). Wait, no. Wait, point C is at (1,√7), and point D is at (-3,√7), so line CD is horizontal at y=√7. Similarly, line AB is horizontal at y=0. Therefore, the distance between AB and CD is √7, which is the height. Therefore, the height h=√7. Therefore, the inradius is h/2=√7 /2. But according to formula r=Area / Perimeter=4√7 /16=√7 /4. Contradiction.Therefore, something is very wrong here. Let me check references.Wait, according to properties of a rhombus: All sides equal, opposite sides parallel, diagonals bisect each other at right angles. A rhombus is a tangential quadrilateral, so it has an incircle. The inradius can be calculated as r= Area / Perimeter. Additionally, in a rhombus, the inradius is also equal to the height of the rhombus divided by 2. Wait, let's confirm.If the height h is the distance between two opposite sides, then Area = base * height = s * h. Therefore, inradius r= Area / Perimeter= (s h)/(4s)= h /4. Wait, so h=4r. Therefore, if h=√7, then r=√7 /4. But according to the distance from center to side, which is h /2=√7 /2. So conflict.Wait, this suggests that the distance from the center to a side is h /2, and since h=4r, then distance from center to side is 2r. But inradius is r, so this is a contradiction. Therefore, clearly, my understanding is flawed.Wait, let me check standard references. For a rhombus with side length a and height h, the area is a*h. The perimeter is 4a. Therefore, inradius r= (a*h)/(4a)=h/4. Therefore, r= h/4.But the distance from the center to a side should be equal to the inradius r. However, according to coordinate calculation, the distance from center to side AB is h/2=√7 /2, which would mean h=√7, so r=√7 /4. Therefore, the formula is correct. Then, why is the distance from center to side AB equal to h/2? Because the center is located halfway between the two opposite sides. So if the distance between AB and CD is h, then the distance from center to AB is h/2. But if inradius is r= h/4, then that suggests that h/2= r*2. Therefore, the distance from the center to the side is twice the inradius. That can't be. Therefore, confusion.Wait, no, no. Wait, inradius is the radius of the incircle, which touches all four sides. Therefore, the distance from the center to each side is equal to the inradius. Therefore, inradius r must be equal to the distance from the center to any side. Therefore, according to coordinate calculation, r=√7 /2. But according to formula r= h /4=√7 /4. Therefore, contradiction. Therefore, one of the premises is wrong.Wait, maybe my coordinate system is not a rhombus. Wait, no, we checked all sides are 4. So it is a rhombus. Then, this suggests that either the formula is wrong, or the coordinate calculation is wrong. But both can't be. Therefore, I must have misunderstood the inradius formula.Wait, let me verify with a square. For a square with side length a, perimeter 4a, area a². Inradius r= a² /4a= a/2. Which is correct, as the inradius is half the side length, which is the distance from center to side. Which is indeed a/2. Therefore, formula works. But in our case, with the rhombus, it's giving r=√7 /4, but coordinate calculation says r=√7 /2. So which is correct.Wait, in the square, the distance from center to side is a/2, which is equal to inradius. Similarly, in our rhombus, distance from center to side is √7 /2, so inradius should be √7 /2. But according to the formula, it's √7 /4. Therefore, discrepancy. Therefore, maybe my formula is wrong.Wait, no. Wait, in a square, which is a rhombus with right angles, the inradius is a/2. But according to the formula for rhombus inradius, which is Area / Perimeter, for square it's (a²)/(4a)=a/4. Wait, wait, that's not correct. Wait, no, no. Wait, inradius for a square is a/2, but according to formula Area / Perimeter: (a²)/(4a)=a/4. But that's not correct. Therefore, my mistake is somewhere else.Wait, no, wait. In a square, the inradius is equal to half the side length. Wait, but the formula Area / Perimeter would be (a²)/(4a)=a/4. Which contradicts. Therefore, this suggests that my formula is wrong. Wait, but inradius is supposed to be the radius of the incircle. For a square, the incircle radius is a/2, but according to Area / Perimeter, it's a² /4a= a/4. Therefore, that's incorrect. Therefore, my mistake is in the formula. Therefore, inradius is not Area / Perimeter for a rhombus. Wait, but according to the general formula for tangential quadrilaterals, inradius is Area / Perimeter.Wait, but in a square, which is both a rhombus and a tangential quadrilateral, the inradius is a/2. But according to Area / Perimeter, (a²)/(4a)=a/4. Therefore, discrepancy. Therefore, the formula inradius=Area / Perimeter must be wrong.Wait, no, wait. Wait, for a square, the area is a², the perimeter is 4a, so inradius is a² /4a= a/4. But in reality, the inradius of a square is a/2. Therefore, this suggests that the formula is incorrect. But that can't be. Wait, no, actually, in a square, the radius of the incircle is a/2, which is indeed equal to Area / Perimeter.Wait, hold on:Wait, area of square is a².Perimeter is 4a.Inradius is a/2.Then, Area / Perimeter = a² /4a =a/4. Which is not equal to the inradius. Therefore, formula inradius=Area / Perimeter is wrong. But that contradicts the general statement for tangential quadrilaterals. Wait, but according to references, inradius for a tangential quadrilateral is indeed Area / semiperimeter. Wait, hold on. Wait, maybe I confused the formula.Wait, no, let me check:For a tangential quadrilateral (one with an incircle), the area is equal to the inradius multiplied by the semiperimeter. So Area= r * semiperimeter. Therefore, r= Area / semiperimeter.Ah! Yes, semiperimeter. Not perimeter. Therefore, correct formula is r=Area / semiperimeter.In the case of the square, semiperimeter is (4a)/2=2a. Therefore, r= a² /2a= a/2, which is correct.Similarly, for the rhombus, semiperimeter is (4s)/2=2s. Therefore, r= Area / (2s). So in our case, Area=4√7, semiperimeter=8, so r=4√7 /8=√7 /2≈1.322. Which matches the coordinate calculation. So my mistake was using perimeter instead of semiperimeter. So the correct formula for inradius is Area / semiperimeter. Therefore, inradius is √7 /2.Therefore, my initial error was dividing by perimeter instead of semiperimeter. Therefore, the correct answer is √7 /2.But let me confirm with our coordinates. The distance from center to side AB is √7 /2, which is indeed the inradius. Therefore, correct answer is √7 /2. Therefore, the previous calculation where I found r=√7 /4 was incorrect because I divided by perimeter instead of semiperimeter.Therefore, the correct inradius is √7 /2.But wait, let's confirm with formula.Inradius r= Area / semiperimeter=4√7 /8=√7 /2. Yes.Therefore, the answer is √7 /2.Therefore, the final answer is boxed{dfrac{sqrt{7}}{2}}.Final Answerboxed{dfrac{sqrt{7}}{2}}</think>Given a rhombus (ABCD) with point (Q) dividing side (BC) in the ratio (1:3) from (B) and point (E) as the midpoint of (AB). It is known that the median (CF) of triangle (CEQ) is (2sqrt{2}) and (EQ = sqrt{2}). We need to find the radius of the circle inscribed in the rhombus.1. Coordinates Setup: - Place (A) at ((0, 0)), (B) at ((s, 0)), (D) at ((p, q)), and (C) at ((s + p, q)). - Midpoint (E) of (AB) is ((s/2, 0)). - Point (Q) divides (BC) in the ratio (1:3), giving coordinates ((s + p/4, q/4)).2. Equations for CF and EQ: - Calculate coordinates of (F) (midpoint of (EQ)): ((frac{3s}{4} + frac{p}{8}, frac{q}{8})). - Use distance formula for (CF) and (EQ) to set up equations: - ( sqrt{left(-frac{s}{4} - frac{7p}{8}right)^2 + left(-frac{7q}{8}right)^2} = 2sqrt{2} ) - ( sqrt{left(frac{s}{2} + frac{p}{4}right)^2 + left(frac{q}{4}right)^2} = sqrt{2} ) - Solve these equations along with (p^2 + q^2 = s^2).3. Solving the Equations: - Substitute (q^2 = s^2 - p^2) into the equations. - Solve for (s), (p), and (q): - (s = 4), (p = -3), (q = sqrt{7}).4. Area and Inradius Calculation: - Calculate the area of the rhombus using the shoelace formula or cross product: (4sqrt{7}). - Calculate the semiperimeter: (2s = 8). - Inradius (r = frac{text{Area}}{text{Semiperimeter}} = frac{4sqrt{7}}{8} = frac{sqrt{7}}{2}).Thus, the radius of the circle inscribed in rhombus (ABCD) is (boxed{dfrac{sqrt{7}}{2}}).]

🔑:Use the formula for the median of a triangle:m^{2}=frac{1}{4}left(2 a^{2}+2 b^{2}-c^{2}right)where a, b, c are the sides of the triangle, and m is the median drawn to the side equal to c.## SolutionWe will use the formula for the median of a triangle:m^{2}=frac{1}{4}left(2 a^{2}+2 b^{2}-c^{2}right)where a, b, c are the sides of the triangle, and m is the median drawn to the side equal to c.Let B Q=x, A C=2 y. Let O be the intersection point of the diagonals of the rhombus A B C D, and M be the midpoint of side B C. ThenO C=y, A B=B C=4 x, M Q=xand since M E is the midline of triangle A B C, we have M E=frac{1}{2} A C=y.Since E Q is the median of triangle B E M, we haveE Q^{2}=frac{1}{4}left(2 B E^{2}+2 E M^{2}-B M^{2}right) text {, or } 2=frac{1}{4}left(2 cdot 4 x^{2}+2 y^{2}-4 x^{2}right)or2 x^{2}+y^{2}=4Since C E is the median of triangle A B C, we haveC E^{2}=frac{1}{4}left(2 A C^{2}+2 B C^{2}-A B^{2}right)=frac{1}{4}left(8 y^{2}+32 x^{2}-16 x^{2}right)=4 x^{2}+2 y^{2}Since C F is the median of triangle C E Q, we have8=C F^{2}=frac{1}{4}left(2 C E^{2}+2 C Q^{2}-E Q^{2}right)=frac{1}{4}left(8 x^{2}+4 y^{2}+18 x^{2}-2right)=frac{1}{4}left(26 x^{2}+4 y^{2}-2right)from which13 x^{2}+y^{2}=17From the obtained system of equationsleft{begin{array}{l}2 x^{2}+y^{2}=4 13 x^{2}+y^{2}=17end{array}right.we find that x=1, y=sqrt{2}. ThenO B^{2}=B C^{2}-O C^{2}=16-2=14Let O P be the height of the right triangle B O C. Then O P is the radius of the circle inscribed in the rhombus. Since B C cdot O P=O B cdot O C, we haveO P=frac{O B cdot O C}{B C}=frac{sqrt{14} cdot sqrt{2}}{4}=frac{sqrt{7}}{2} text {. }Answerfrac{sqrt{7}}{2}.