🧠:[<think>Okay, so I need to find the minimum value of this function f(x) which is the sum of three square roots. Each of those square roots has trigonometric functions inside, so this seems like it might involve some trigonometric identities or maybe calculus to find the minimum. Let me break it down step by step.First, let's write down the function again to make sure I have it right:f(x) = √(10 - 6 cos x) + √(17/8 - (3√2)/2 sin x) + √(19 - 2√2 cos x - 8 sin x)Hmm, each term under the square roots is a combination of sine and cosine functions. Since the function is periodic (because sine and cosine are periodic), the minimum should occur at some point within the interval [0, 2π). Maybe I can use calculus here. To find the minimum, I need to take the derivative of f(x) with respect to x, set it equal to zero, and solve for x. But the function is a sum of square roots, so the derivatives might get complicated. Let me think.Alternatively, maybe there's a geometric interpretation here. Sometimes expressions involving √(A - B cos x - C sin x) can represent distances in coordinate systems. For example, if I think of points in the plane, the distance from a moving point on a circle to another fixed point might give rise to such expressions. Let me explore that idea.Let me recall that the expression √(D - E cos x - F sin x) can be rewritten using the formula for the distance between two points. If we have a point (a, b) and another point (c, d), the distance between them is √[(a - c)^2 + (b - d)^2]. If one of the points is parameterized by an angle x, say (G cos x, G sin x), then the distance becomes √[(G cos x - c)^2 + (G sin x - d)^2]. Expanding that would give √[G² + c² + d² - 2G c cos x - 2G d sin x]. Comparing this to the terms in the square roots of f(x), maybe each square root term represents a distance from a point on a circle to another fixed point.Let me check each term:1. First term: √(10 - 6 cos x)Let me see if this can be written as a distance. Let's suppose there is a point moving on a circle with radius r centered at the origin, so coordinates (r cos x, r sin x). The distance from this point to a fixed point (a, b) would be √[(r cos x - a)^2 + (r sin x - b)^2]. Expanding this gives √[r² + a² + b² - 2 r a cos x - 2 r b sin x]. Comparing to the first term √(10 - 6 cos x). So if I set the coefficient of sin x to zero, which would mean 2 r b = 0, so either r=0 or b=0. But r can't be zero because we have a radius. So b must be 0. Then the expression under the square root becomes r² + a² - 2 r a cos x. Comparing to 10 -6 cos x. Therefore:r² + a² = 10and2 r a = 6 ⇒ r a = 3So we have two equations:r² + a² = 10andr a = 3We can solve for a in terms of r: a = 3/r. Substitute into the first equation:r² + (9)/(r²) = 10Multiply both sides by r²:r⁴ + 9 = 10 r²Which is r⁴ -10 r² +9=0. Let’s set y = r²:y² -10 y +9=0 ⇒ y = [10 ± √(100 -36)]/2 = [10 ± √64]/2 = [10 ±8]/2 ⇒ y= (18)/2=9 or y=(2)/2=1. Therefore, r²=9 ⇒ r=3 or r²=1 ⇒ r=1. Then a=3/r: if r=3, then a=1; if r=1, then a=3. So possible points: If the moving point is on a circle of radius 3, then the fixed point is (1,0). Alternatively, if the moving point is on a circle of radius 1, the fixed point is (3,0). But since the expression is √(10 -6 cos x), which is the same as the distance between (3 cos x, 3 sin x) and (1,0):Distance squared: (3 cos x -1)^2 + (3 sin x)^2 = 9 cos²x -6 cos x +1 +9 sin²x= 9 (cos²x + sin²x) -6 cos x +1= 9 -6 cos x +1=10 -6 cos x. Exactly. So the first term is the distance between (3 cos x, 3 sin x) and (1,0).Similarly, let's check the second term: √(17/8 - (3√2)/2 sin x). Let me see if this can also be represented as a distance. Let's suppose we have a moving point on a circle with radius s, centered at the origin: (s cos x, s sin x). The distance to a fixed point (c, d):√[(s cos x - c)^2 + (s sin x - d)^2] = √[s² + c² + d² - 2 s c cos x - 2 s d sin x]. Comparing to the second term under the square root: 17/8 - (3√2)/2 sin x. So, we need:s² + c² + d² =17/8and-2 s c cos x term is not present, so coefficient of cos x must be zero: 2 s c =0, so either s=0 or c=0. Since s is a radius, s≠0, so c=0.Then the coefficient of sin x is -2 s d = -(3√2)/2 ⇒ 2 s d = (3√2)/2 ⇒ s d = (3√2)/4.So we have c=0, so fixed point is (0, d). Then:s² + 0 + d² =17/8 ⇒ s² + d² =17/8ands d = (3√2)/4Let me solve these equations. Let’s let’s denote s² + d² =17/8 and s d = (3√2)/4. Let me set variables: Let’s call s d = k = (3√2)/4. Then, note that (s + d)^2 = s² + 2 s d + d² = (17/8) + 2*(3√2)/4 =17/8 + (3√2)/2. Hmm, not sure if that helps. Alternatively, maybe we can express d = k/s. Then substitute into s² + d² =17/8:s² + (k²)/s² =17/8Multiply both sides by s²:s⁴ + k² = (17/8) s²s⁴ - (17/8)s² + k² =0With k = (3√2)/4, so k² = (9*2)/16= 9/8.Thus, the equation becomes:s⁴ - (17/8)s² +9/8=0Multiply through by 8 to eliminate denominators:8 s⁴ -17 s² +9=0Let’s set y=s²:8 y² -17 y +9=0Solutions:y = [17 ±√(289 - 288)]/(16)= [17 ±1]/16So y=(17+1)/16=18/16=9/8, or y=(17-1)/16=16/16=1.Therefore, s²=9/8 or s²=1. Then s= 3/(2√2) or s=1.Then d= k/s= (3√2)/4 divided by s:If s=3/(2√2), then d= (3√2)/4 / (3/(2√2)) )= (3√2)/4 * (2√2)/3 = (2*2)/4=1. So d=1.If s=1, then d= (3√2)/4 /1= (3√2)/4.Therefore, two possibilities:Either the moving point is on a circle of radius 3/(2√2) and the fixed point is (0,1), or the moving point is on a circle of radius 1 and the fixed point is (0, 3√2/4).Let me check which one gives the correct expression.First case: s=3/(2√2), fixed point (0,1). Compute the distance squared:(s cos x -0)^2 + (s sin x -1)^2= s² cos²x + s² sin²x - 2 s sin x +1= s² (cos²x + sin²x) -2 s sin x +1= s² -2 s sin x +1.s²= (9)/(8), so:9/8 -2*(3/(2√2)) sin x +1=9/8 +1 - (3/√2) sin x=17/8 - (3√2)/2 sin x. Wait, 3/√2 is equal to (3√2)/2, right? Because 3/√2 = 3√2/2. Yes. So that gives exactly the expression under the second square root. So the second term is the distance between the point (3/(2√2) cos x, 3/(2√2) sin x) and (0,1).Alternatively, the second case s=1, fixed point (0, 3√2/4). Then distance squared:(cos x)^2 + (sin x - 3√2/4)^2= cos²x + sin²x - (3√2/2) sin x + (9*2)/16=1 - (3√2/2) sin x + 9/8= (8/8 +9/8) - (3√2/2) sin x=17/8 - (3√2/2) sin x. Same result. So either way, the second term is the distance between a point on a circle of radius 3/(2√2) or 1 to (0,1) or (0, 3√2/4). But both interpretations are valid; maybe the first one is more useful.Similarly, let's check the third term: √(19 - 2√2 cos x -8 sin x). Let's see if this can also be a distance. Using the same approach, if the moving point is (t cos x, t sin x), distance to (e, f):√[(t cos x - e)^2 + (t sin x -f)^2] = √[t² + e² + f² -2 t e cos x -2 t f sin x]Comparing to 19 -2√2 cos x -8 sin x. So:t² + e² + f² =19and2 t e =2√2 ⇒ t e=√22 t f=8 ⇒ t f=4So we have equations:t e =√2t f=4And t² + e² + f²=19Let me solve for e and f in terms of t:From t e=√2 ⇒ e=√2 / tFrom t f=4 ⇒ f=4/tSubstitute into the third equation:t² + (√2 / t)^2 + (4/t)^2 =19Compute:t² + (2)/t² + 16/t²= t² + 18/t² =19Multiply both sides by t²:t⁴ +18=19 t² ⇒ t⁴ -19 t² +18=0Let y =t²:y² -19 y +18=0 ⇒ y=(19 ±√(361 -72))/2=(19 ±√289)/2=(19 ±17)/2Thus, y=(19+17)/2=36/2=18 or y=(19-17)/2=2/2=1. So t²=18 or t²=1. Thus, t=√18=3√2 or t=1.If t=3√2, then e=√2/(3√2)=1/3, and f=4/(3√2)= (4√2)/(6)= (2√2)/3.If t=1, then e=√2/1=√2, and f=4/1=4.Therefore, the third term can be interpreted as the distance between the point (3√2 cos x, 3√2 sin x) and (1/3, 2√2/3) or between (cos x, sin x) and (√2,4). Let me check which one.First case: t=3√2, fixed point (1/3, 2√2/3). Compute distance squared:(3√2 cos x -1/3)^2 + (3√2 sin x -2√2/3)^2Expand:(9*2 cos²x - 2*(3√2)(1/3) cos x +1/9) + (9*2 sin²x - 2*(3√2)(2√2/3) sin x + (8/9))Simplify term by term:First term: 18 cos²x - 2*(√2) cos x +1/9Second term: 18 sin²x - 2*(3√2)*(2√2)/3 sin x +8/9Simplify coefficients:-2*(3√2)*(2√2)/3 = -2*(2*2) = -8So second term becomes 18 sin²x -8 sin x +8/9Combine both terms:18 cos²x +18 sin²x -2√2 cos x -8 sin x +1/9 +8/9Simplify:18 (cos²x + sin²x) -2√2 cos x -8 sin x + (1+8)/9=18(1) -2√2 cos x -8 sin x +1=18 +1 -2√2 cos x -8 sin x=19 -2√2 cos x -8 sin xPerfect, so the third term is the distance between (3√2 cos x,3√2 sin x) and (1/3, 2√2/3).Alternatively, the other case: t=1, fixed point (√2,4). Compute distance squared:(cos x -√2)^2 + (sin x -4)^2= cos²x -2√2 cos x +2 + sin²x -8 sin x +16= (cos²x + sin²x) + (-2√2 cos x -8 sin x) +18=1 -2√2 cos x -8 sin x +18=19 -2√2 cos x -8 sin x. Same result. So either interpretation is valid. But maybe the first one with t=3√2 is more useful.So now, putting this all together, the function f(x) is the sum of three distances:1. From (3 cos x, 3 sin x) to (1,0)2. From ( (3/(2√2)) cos x, (3/(2√2)) sin x ) to (0,1)3. From (3√2 cos x, 3√2 sin x ) to (1/3, 2√2/3)But wait, hold on. The moving points for each term are different: each term has a different radius. The first term's moving point is (3 cos x, 3 sin x), the second is ( (3/(2√2)) cos x, (3/(2√2)) sin x ), and the third is (3√2 cos x, 3√2 sin x ). These are three different points moving on three different circles as x varies. But x is the same variable for all three terms, so all three moving points are parameterized by the same angle x but with different radii. That complicates things because if we try to interpret f(x) as the sum of three distances from points moving on different circles to their respective fixed points, all parameterized by the same angle x. But since each term has a different radius, they don't all correspond to the same point moving on a circle. So maybe this geometric interpretation isn't straightforward. Hmm.Alternatively, perhaps there's a way to scale each term so that all moving points have the same radius. Let me check:First term: radius 3Second term: radius 3/(2√2) ≈ 3/(2.828) ≈1.06Third term: radius 3√2≈4.24So they are all different. Hmm. So perhaps this approach isn't directly helpful. Alternatively, maybe there's a substitution for angle x. Let me think.Alternatively, perhaps we can use the Cauchy-Schwarz inequality or some other inequality to find the minimum. But since f(x) is the sum of three square roots, each involving trigonometric functions, it's not obvious how to apply inequalities here. Alternatively, maybe we can parameterize x in terms of sine and cosine, using a single parameter, like t=tan(x/2), but that might complicate the expressions further.Alternatively, consider using calculus. Let's denote the three terms as:A(x) = √(10 -6 cos x)B(x) = √(17/8 - (3√2)/2 sin x)C(x) = √(19 -2√2 cos x -8 sin x)Then f(x)=A(x)+B(x)+C(x). To find the minimum, compute f'(x) and set to zero.Compute derivatives:A’(x)= (1/(2√(10 -6 cos x)))*(6 sin x)= (3 sin x)/√(10 -6 cos x)Similarly,B’(x)= (1/(2√(17/8 - (3√2)/2 sin x)))*( - (3√2)/2 cos x )= (-3√2 cos x)/(2 √(17/8 - (3√2)/2 sin x))C’(x)= (1/(2√(19 -2√2 cos x -8 sin x)))*(2√2 sin x -8 cos x)= (2√2 sin x -8 cos x)/(2 √(19 -2√2 cos x -8 sin x))Thus, f’(x)= A’(x)+B’(x)+C’(x)=0.Setting this equal to zero:(3 sin x)/√(10 -6 cos x) - (3√2 cos x)/(2 √(17/8 - (3√2)/2 sin x)) + (2√2 sin x -8 cos x)/(2 √(19 -2√2 cos x -8 sin x))=0This looks very complicated. Solving this equation analytically seems difficult. Maybe we can look for critical points by testing specific angles x where the trigonometric functions take simple values, like x=0, π/2, π, 3π/2, etc., or angles where the expressions under the square roots become perfect squares, which might simplify the derivatives.Let me try evaluating f(x) at x=0, π/2, π, 3π/2, etc.First, x=0:Compute each term:A(0)=√(10 -6 cos0)=√(10 -6*1)=√4=2B(0)=√(17/8 - (3√2)/2 sin0)=√(17/8 -0)=√(17/8)=√(17)/2√2≈2.06/2.828≈0.73C(0)=√(19 -2√2 cos0 -8 sin0)=√(19 -2√2*1 -0)=√(19 -2√2)≈√(19 -2.828)≈√16.172≈4.021Thus f(0)=2 +0.73 +4.021≈6.751x=π/2:A(π/2)=√(10 -6 cos π/2)=√10≈3.162B(π/2)=√(17/8 - (3√2)/2 sin π/2)=√(17/8 - (3√2)/2*1)=√(17/8 -3√2/2). Let's compute 17/8 ≈2.125, 3√2/2≈2.121, so 2.125 -2.121≈0.004. So B(π/2)=√0.004≈0.063C(π/2)=√(19 -2√2 cos π/2 -8 sin π/2)=√(19 -0 -8*1)=√11≈3.316Thus f(π/2)=3.162 +0.063 +3.316≈6.541x=π:A(π)=√(10 -6 cos π)=√(10 -6*(-1))=√16=4B(π)=√(17/8 - (3√2)/2 sin π)=√(17/8 -0)=√(17/8)=≈1.46C(π)=√(19 -2√2 cos π -8 sin π)=√(19 -2√2*(-1) -0)=√(19 +2√2)≈√(19 +2.828)=√21.828≈4.673Thus f(π)=4 +1.46 +4.673≈10.133x=3π/2:A(3π/2)=√(10 -6 cos 3π/2)=√10≈3.162B(3π/2)=√(17/8 - (3√2)/2 sin3π/2)=√(17/8 - (3√2)/2*(-1))=√(17/8 +3√2/2). Let's compute 17/8≈2.125, 3√2/2≈2.121, so sum≈4.246. So √4.246≈2.06C(3π/2)=√(19 -2√2 cos3π/2 -8 sin3π/2)=√(19 -0 -8*(-1))=√(19 +8)=√27≈5.196Thus f(3π/2)=3.162 +2.06 +5.196≈10.418So among these, x=π/2 gives the smallest value so far, around 6.541. Let's check some other angles. Maybe x=π/4 or 3π/4.x=π/4:Compute each term:cos π/4=√2/2≈0.707, sin π/4=√2/2≈0.707A(π/4)=√(10 -6*(√2/2))=√(10 -3√2)≈√(10 -4.242)=√5.758≈2.398B(π/4)=√(17/8 - (3√2)/2*(√2/2))=√(17/8 - (3*2)/4)=√(17/8 -3/2)=√(17/8 -12/8)=√(5/8)≈0.7906C(π/4)=√(19 -2√2*(√2/2) -8*(√2/2))=√(19 - (2*2)/2 -4√2)=√(19 -2 -5.656)=√11.344≈3.368Thus f(π/4)=2.398 +0.7906 +3.368≈6.556Hmm, slightly higher than at x=π/2.x=3π/4:cos3π/4=-√2/2≈-0.707, sin3π/4=√2/2≈0.707A(3π/4)=√(10 -6*(-√2/2))=√(10 +3√2)≈√(10 +4.242)=√14.242≈3.774B(3π/4)=√(17/8 - (3√2)/2*(√2/2))= same as π/4: √(5/8)≈0.7906C(3π/4)=√(19 -2√2*(-√2/2) -8*(√2/2))=√(19 + (2*2)/2 -4√2)=√(19 +2 -5.656)=√15.344≈3.917Thus f(3π/4)=3.774 +0.7906 +3.917≈8.481Okay, still higher than π/2.What about x=π/6?x=π/6: cos π/6=√3/2≈0.866, sin π/6=0.5A(π/6)=√(10 -6*(√3/2))=√(10 -3√3)≈√(10 -5.196)=√4.804≈2.192B(π/6)=√(17/8 - (3√2)/2*(0.5))=√(17/8 - (3√2)/4)≈√(2.125 -1.06)=√1.065≈1.032C(π/6)=√(19 -2√2*(√3/2) -8*(0.5))=√(19 -√6 -4)=√(15 -2.449)=√12.551≈3.543f(π/6)=2.192 +1.032 +3.543≈6.767Still higher than x=π/2.What about x=π/3?cos π/3=0.5, sin π/3≈0.866A(π/3)=√(10 -6*0.5)=√(10 -3)=√7≈2.645B(π/3)=√(17/8 - (3√2)/2*(√3/2))=√(17/8 - (3√6)/4)≈√(2.125 -1.837)=√0.288≈0.537C(π/3)=√(19 -2√2*0.5 -8*(√3/2))=√(19 -√2 -4√3)≈√(19 -1.414 -6.928)=√10.658≈3.266Thus f(π/3)=2.645 +0.537 +3.266≈6.448That's lower than at x=π/2 (≈6.541). Wait, really?Wait, let's compute more accurately:At x=π/3:A(π/3)=√7≈2.6458B(π/3)=√(17/8 - (3√2)/2*(√3/2))=√(17/8 - (3√6)/4). Let's compute exact value:17/8=2.1253√6/4≈3*2.449/4≈7.347/4≈1.8368Thus, 2.125 -1.8368≈0.2882. So sqrt(0.2882)=≈0.5367C(π/3)=√(19 -2√2*(1/2) -8*(√3/2))=√(19 -√2 -4√3). Compute:√2≈1.414, 4√3≈6.928So 19 -1.414 -6.928≈19 -8.342≈10.658. sqrt(10.658)=≈3.266Thus total≈2.6458 +0.5367 +3.266≈6.448. Which is lower than x=π/2's 6.541. So x=π/3 might be better.Hmm, so maybe π/3 is a better candidate.Let me check x=π/3 more carefully. Also, maybe x=π/6, but previous result was higher.Alternatively, perhaps the minimum is near x=π/3. Let's try x=π/3.5 or something. Wait, maybe we can use calculus here. But since the derivative is complicated, perhaps we can use numerical methods to approximate the minimum.Alternatively, note that at x=π/4, the value was≈6.556, at x=π/3≈6.448, at x=π/2≈6.541, so perhaps the minimum is around x=π/3. Let me check x=5π/12 (75 degrees).x=5π/12. Let's compute:cos5π/12≈cos75°≈0.2588, sin5π/12≈0.9659.Compute each term:A=√(10 -6*0.2588)=√(10 -1.5528)=√8.4472≈2.907B=√(17/8 - (3√2)/2*0.9659). Compute (3√2)/2*0.9659≈3*1.414/2*0.9659≈2.121*0.9659≈2.05. Then 17/8≈2.125, so 2.125 -2.05≈0.075. sqrt≈0.274.C=√(19 -2√2*0.2588 -8*0.9659). Compute:2√2*0.2588≈2*1.414*0.2588≈2.828*0.2588≈0.7318*0.9659≈7.72719 -0.731 -7.727≈10.542. sqrt≈3.247Thus total≈2.907 +0.274 +3.247≈6.428. So lower than π/3.Hmm, so even lower. Wait, maybe even lower. Let me check x= π/3.5? Not sure. Alternatively, maybe x=π/4.5? But this is getting messy.Alternatively, perhaps the minimum occurs when all three terms are minimized simultaneously? But likely they are not minimized at the same x. Alternatively, perhaps there's a geometric configuration where the three distances add up to a minimum. Since each term is a distance from a point on a circle to a fixed point, maybe the sum is minimized when the three moving points are colinear with their respective fixed points in some coordinated way. But this is vague.Alternatively, going back to the geometric interpretation, but considering that each term is a distance from a different point on a different circle (with radii 3, 3/(2√2), and 3√2) all parameterized by the same angle x. So if we imagine three different circles, each with their own radii, and a point on each circle all rotating with angle x, and we are measuring the distance from each point to their respective fixed points, then summing these distances. The problem is that since the radii are different, the points are moving at different "speeds" as x changes, so it's not straightforward to model this as a single point moving in some composite space.Alternatively, perhaps we can parametrize all three terms using a common parameter. Let me consider using complex numbers. Each moving point can be represented as a complex number: 3 e^{ix}, (3/(2√2)) e^{ix}, and 3√2 e^{ix}. Then their respective fixed points are 1, i, and (1/3 + 2√2/3 i). Then f(x) is the sum of the magnitudes of (3 e^{ix} -1), ((3/(2√2)) e^{ix} - i), and (3√2 e^{ix} - (1/3 + 2√2/3 i)). Hmm, but how does this help? Maybe by considering vector addition or something? Not sure.Alternatively, maybe using the triangle inequality. But since these are three different vectors, the sum of their magnitudes isn't easily related to the magnitude of their sum. Unless there's a particular alignment where all three vectors are colinear or something, but this seems unlikely.Alternatively, maybe there's a substitution for t = sin x or cos x. Let me try to set t = sin x, then cos x = √(1 - t²), but this introduces square roots and complicates differentiation. Alternatively, use a substitution like t = tan(x/2), which can parameterize sin x and cos x, but that might lead to messy expressions.Alternatively, consider combining the terms. For example, perhaps write the three terms as vectors and find when their sum has minimal magnitude? But again, not sure.Alternatively, note that each term is a distance, so maybe the problem is equivalent to finding the minimal total distance from three different points moving on three circles to their respective fixed points. This is similar to a Fermat-Toricelli problem but with moving points. However, such problems generally don't have analytical solutions and require numerical optimization.Alternatively, maybe there's a clever trigonometric identity that can combine these terms. Let me look at the expressions under the square roots:First term: 10 -6 cos xSecond term:17/8 - (3√2)/2 sin xThird term:19 -2√2 cos x -8 sin xPerhaps we can express these in terms of a single angle or combine them into a single expression. Let's see:Suppose we consider the third term:19 -2√2 cos x -8 sin x. Let me try to write this as A - B cos x - C sin x, and then express it as D - E cos(x + φ) using the identity B cos x + C sin x = E cos(x + φ). Let's do that for each term.Starting with the first term:10 -6 cos x. This is already in the form D - E cos x, so no need to adjust.Second term:17/8 - (3√2)/2 sin x. Similarly, this is D - C sin x. We can write this as D - C sin x = D - C cos(x - π/2), since sin x = cos(x - π/2). So 17/8 - (3√2)/2 cos(x - π/2). Then using the same identity, this can be written as D - E cos(x - π/2 + φ), but maybe this is not helpful.Third term:19 -2√2 cos x -8 sin x. Let me combine the cosine and sine terms:-2√2 cos x -8 sin x = - [2√2 cos x +8 sin x]. Let me write this as -R cos(x - φ), where R=√[(2√2)^2 +8^2]=√[8 +64]=√72=6√2. Then:2√2 cos x +8 sin x=6√2 cos(x - φ), where φ=arctan(8/(2√2))=arctan(4/√2)=arctan(2√2). So φ=arctan(2√2). Therefore, the third term becomes 19 -6√2 cos(x - φ). So √(19 -6√2 cos(x - φ)). Similarly, maybe the other terms can be written in similar forms. But does this help? Perhaps, if the angles φ are related in such a way that the cosines can be aligned.Alternatively, note that the third term is similar to the first term. The first term is √(10 -6 cos x), which can be written as √(10 -6 cos x). If we can write all three terms as √(A - B cos(x - θ)) for some angles θ, then perhaps we can use some identity or optimization over θ. But this seems a stretch.Alternatively, perhaps the minimal value occurs when all the cosine and sine terms are minimized such that each square root term is minimized. But the terms are interdependent because the angle x is the same for all. Minimizing one term might increase another. For instance, when x=π/2, the first term is √10≈3.16, the second term is minimized (≈0.063), and the third term is √11≈3.32. But when x=0, the first term is minimized (2), but the second term is larger (≈0.73), and the third term is≈4.02. The sum at x=0 is≈6.75, which is higher than at x=π/2 (≈6.54). So the minimal sum doesn't necessarily occur at the minimal of individual terms.Alternatively, maybe there's a point where the derivative is zero, which requires a balance between the rates of change of each term. Given the complexity of the derivative, maybe we can set up an equation where the sum of the derivatives equals zero and attempt to solve it numerically.Alternatively, maybe we can consider specific relations between the coefficients. Let me look at the third term again:19 -2√2 cos x -8 sin x. Suppose we factor out something:Let me try to write -2√2 cos x -8 sin x as a multiple of a single sine or cosine function. As before:-2√2 cos x -8 sin x = -R cos(x + φ), where R=√[(2√2)^2 +8^2]=√[8 +64]=√72=6√2. Then:R cos(x + φ)=2√2 cos x +8 sin x. So:cos(x + φ)= (2√2)/R cos x +8/R sin x= (2√2)/(6√2) cos x +8/(6√2) sin x= (1/3) cos x + (4)/(3√2) sin xThus, φ=arctan( (4/(3√2)) / (1/3) )=arctan(4/√2)=arctan(2√2). Therefore, the third term can be written as √(19 -6√2 cos(x + arctan(2√2)) )But I don't see how this helps directly. Similarly, the first term is √(10 -6 cos x), the second term is √(17/8 - (3√2)/2 sin x), and the third term is √(19 -6√2 cos(x + φ)). Still not obvious.Alternatively, think of all three terms as projections or related to vectors. Maybe use the Cauchy-Schwarz inequality. But I need to recall that for vectors u and v, |u + v| ≤ |u| + |v|. But here, we have three terms added as scalars, not vectors.Alternatively, consider the function f(x) as the sum of three distances. If we can interpret these distances geometrically as sides of a triangle or something, but again, not straightforward.Wait, perhaps the key is that all three terms can be considered as distances from points on different circles parameterized by the same angle x to fixed points. If we can find an x such that all three moving points are aligned in a way that minimizes the total distance. This is similar to a traveling salesman problem but in a geometric setting.Alternatively, since all three moving points are parameterized by the same angle x, maybe there's a way to represent this as a single point in a higher-dimensional space, but that might be overcomplicating.Alternatively, notice that the coefficients in each term might be related. Let me check:First term: 10 -6 cos x. The coefficients are 10 and 6.Second term:17/8 - (3√2)/2 sin x. Coefficients are 17/8 and (3√2)/2≈2.121.Third term:19 -2√2 cos x -8 sin x. Coefficients are 19, 2√2≈2.828, and 8.Is there any relation between 10, 17/8, and 19? Not obvious. 10 +17/8 +19=10+2.125+19=31.125. Not helpful.Alternatively, check if there's a common multiple or ratio. For example, 6, 3√2/2, 2√2. Let's see:6, (3√2)/2, 2√2. Let's express all in terms of √2:6=6/1≈6(3√2)/2≈2.1212√2≈2.828Not obvious.Alternatively, think of 6=3*2, 3√2/2= 3/2*√2, 2√2=2*√2. Maybe there's a common factor of √2 in some terms.Alternatively, maybe consider the following: If I can write all three terms as distances, then the sum of distances might be minimized when the moving points are such that the angles are arranged to make the path from the fixed points to the moving points colinear or something. But since each moving point is on a different circle, this seems difficult.Alternatively, since the problem is asking for the minimal value, maybe there's a lower bound that can be found using the Cauchy-Schwarz inequality or the Minkowski inequality (triangle inequality for sums).Minkowski inequality states that for vectors a, b, c, ||a + b + c|| ≤ ||a|| + ||b|| + ||c||. But here, we have scalar distances, not vectors. So not directly applicable.Alternatively, think of each term as a norm and apply Hölder's inequality. But Hölder's inequality relates sums and products, which might not be directly useful here.Alternatively, use the Cauchy-Schwarz inequality on the sum:f(x)=√(10 -6 cos x)+√(17/8 - (3√2)/2 sin x)+√(19 -2√2 cos x -8 sin x)Suppose I consider each square root as the dot product of some vectors. Wait, not sure.Alternatively, note that each term under the square root is of the form A - B cos x - C sin x. Since cos x and sin x can be represented as components of a unit vector, maybe each term is related to the projection of some vector onto the direction x. If so, then the sum might have an interpretation as the total projection, but I don't see how this leads to a minimum.Alternatively, consider that for expressions like A - B cos x - C sin x, the minimum value of the square root would be √(A - √(B² + C²)), achieved when cos x = B / √(B² + C²) and sin x = C / √(B² + C²). But since we are dealing with the sum of three such terms, which each have their own A, B, C, and the angle x must be the same for all, it's unclear if they can all achieve their individual minima simultaneously.Let me check if the angles required to minimize each term individually are compatible.First term: √(10 -6 cos x). Minimum when cos x=1, x=0. Minimum value √4=2.Second term: √(17/8 - (3√2)/2 sin x). Minimum when sin x=1, x=π/2. Minimum value √(17/8 -3√2 / 2). As before, ≈√0.004≈0.063.Third term: √(19 -2√2 cos x -8 sin x). Let's find its minimum. The expression under the root is 19 -2√2 cos x -8 sin x. To minimize this, we need to maximize 2√2 cos x +8 sin x. The maximum value of B cos x + C sin x is √(B² + C²). So here, √( (2√2)^2 +8^2 )=√(8 +64)=√72=6√2. Therefore, the minimum of the third term is √(19 -6√2). Compute 6√2≈8.485, so 19 -8.485≈10.515. Therefore, √10.515≈3.242. This minimum occurs when cos x= (2√2)/√72= (2√2)/(6√2)=1/3, and sin x=8/√72=8/(6√2)=4/(3√2)=2√2/3. Therefore, x=arcsin(2√2/3). Let me compute that angle. sin x=2√2/3≈0.9428. Wait, 2√2≈2.828, 2.828/3≈0.9428, which is greater than 1. But wait, that's impossible. Wait, no: √72=6√2≈8.485. So B=2√2, C=8. Then the maximum of B cos x + C sin x is √( (2√2)^2 +8^2 )=√(8+64)=√72=6√2≈8.485. But 19 -8.485≈10.515, so √10.515≈3.242. However, to achieve that maximum, we need cos x= B / √(B² + C²)=2√2 /6√2=1/3, and sin x=8 /6√2=4/(3√2)= (4√2)/6= (2√2)/3≈0.9428. Wait, but 2√2/3≈0.9428, which is less than 1 (since 2√2≈2.828, 2.828/3≈0.9428). So x=arcsin(2√2/3)≈arcsin(0.9428)≈70.5 degrees. So x≈1.23 radians. Therefore, the third term's minimum occurs around x≈1.23. Let's check if this x also minimizes the sum f(x).At x≈1.23 radians (≈70.5 degrees):Compute each term:First term: √(10 -6 cos x). cos(1.23)≈cos70.5≈0.33, so 10 -6*0.33≈10 -1.98≈8.02. sqrt≈2.83.Second term: √(17/8 - (3√2)/2 sin x). sin(1.23)≈sin70.5≈0.9428. Then (3√2)/2 *0.9428≈2.121 *0.9428≈2.0. So 17/8 -2.0≈2.125 -2.0=0.125. sqrt≈0.353.Third term: √(19 -2√2 cos x -8 sin x). As computed,≈3.242.Thus total≈2.83 +0.353 +3.242≈6.425.This is lower than previous values. So perhaps the minimum is around 6.42. Let's check at x=1.2 radians:x=1.2 radians≈68.8 degrees.cos1.2≈0.362, sin1.2≈0.932.First term:√(10 -6*0.362)=√(10 -2.172)=√7.828≈2.798Second term:√(17/8 - (3√2)/2*0.932). Compute (3√2)/2≈2.121, 2.121*0.932≈1.976. Then 17/8≈2.125 -1.976≈0.149. sqrt≈0.386Third term:√(19 -2√2*0.362 -8*0.932). Compute 2√2*0.362≈2.828*0.362≈1.024, 8*0.932≈7.456. So 19 -1.024 -7.456≈10.52. sqrt≈3.244Total≈2.798 +0.386 +3.244≈6.428Close to the previous value. Let me check x=1.25 radians≈71.6 degrees.cos1.25≈0.315, sin1.25≈0.949.First term:√(10 -6*0.315)=√(10 -1.89)=√8.11≈2.848Second term:√(17/8 - (3√2)/2 *0.949). (3√2)/2≈2.121*0.949≈2.013. 17/8 -2.013≈2.125 -2.013≈0.112. sqrt≈0.335Third term:√(19 -2√2*0.315 -8*0.949). 2√2*0.315≈2.828*0.315≈0.891. 8*0.949≈7.592. 19 -0.891 -7.592≈10.517. sqrt≈3.242Total≈2.848 +0.335 +3.242≈6.425Same as before. It seems like the minimum is around 6.42-6.43. However, since the problem is likely expecting an exact value, maybe there's a way to find an exact angle where the sum is minimized. Alternatively, perhaps the minimal value is achieved when all three terms are minimized in some coordinated way.Wait, looking back at the third term's minimum condition, where x≈1.23 radians, which corresponds to sin x=2√2/3. Let's see if this angle also minimizes the sum. Let's compute f(x) at sin x=2√2/3 and cos x=1/3.Wait, from the third term's optimization, we have:cos x=1/3 and sin x=2√2/3. Let me verify this:(1/3)^2 + (2√2/3)^2=1/9 +8/9=9/9=1. Yes, so this is a valid angle x where cos x=1/3 and sin x=2√2/3.Let's compute f(x) at this angle.Compute each term:A(x)=√(10 -6 cos x)=√(10 -6*(1/3))=√(10 -2)=√8=2√2≈2.828B(x)=√(17/8 - (3√2)/2 sin x)=√(17/8 - (3√2)/2*(2√2/3))=√(17/8 - (3√2)*(2√2)/6)=√(17/8 - (12/6))=√(17/8 -2)=√(17/8 -16/8)=√(1/8)=1/(2√2)≈0.3535C(x)=√(19 -2√2 cos x -8 sin x)=√(19 -2√2*(1/3) -8*(2√2/3))=√(19 - (2√2)/3 - (16√2)/3)=√(19 - (18√2)/3)=√(19 -6√2)≈√(19 -8.485)≈√10.515≈3.242Thus, f(x)=2√2 +1/(2√2)+√(19 -6√2). Let's compute this exactly:First term:2√2Second term:1/(2√2)=√2/4Third term:√(19 -6√2)So total f(x)=2√2 +√2/4 +√(19 -6√2)Simplify 2√2 +√2/4= (8√2 +√2)/4= (9√2)/4≈ (9*1.414)/4≈12.726/4≈3.1815Add third term≈3.242Total≈3.1815 +3.242≈6.4235Which matches our numerical approximation earlier. So the minimal value is exactly 9√2/4 +√(19 -6√2). But wait, can this expression be simplified further?Let me compute 9√2/4 +√(19 -6√2). Maybe there's a way to rationalize or combine these terms.Alternatively, note that 19 -6√2= (sqrt(2))^0 something? Let me compute:Suppose 19 -6√2 can be expressed as (a - b√2)^2. Let's check:(a -b√2)^2= a² +2b² -2ab√2=19 -6√2. So equate:a² +2b²=192ab=6 ⇒ ab=3So solving:From ab=3, a=3/b. Substitute into first equation:(9/b²) +2b²=19Multiply by b²:9 +2b⁴=19 b²2b⁴ -19 b² +9=0Let y=b²:2y² -19y +9=0Solutions:y=(19 ±√(361 -72))/4=(19 ±√289)/4=(19 ±17)/4Thus, y=(19+17)/4=36/4=9 or y=(19-17)/4=2/4=0.5Thus, b²=9⇒b=3 or b²=0.5⇒b=1/√2Thus, if b=3, then a=3/b=1. Then (1 -3√2)^2=1 +18 -6√2=19 -6√2. Yes! So √(19 -6√2)=|1 -3√2|=3√2 -1, since 3√2≈4.242 >1. So √(19 -6√2)=3√2 -1Therefore, f(x)=9√2/4 +3√2 -1= (9√2/4 +12√2/4) -1=21√2/4 -1But wait, let's check:√(19 -6√2)=3√2 -1. Let's square 3√2 -1:(3√2)^2 -2*3√2*1 +1=18 -6√2 +1=19 -6√2. Exactly. So yes, √(19 -6√2)=3√2 -1.Therefore, f(x)=9√2/4 +3√2 -1= (9√2 +12√2)/4 -1=21√2/4 -1= (21√2 -4)/4.But 21√2 is approximately 29.698, so 29.698 -4=25.698, divided by4≈6.4245, which matches our previous numerical results.Therefore, the minimal value is (21√2 -4)/4. But let's check if this can be simplified further or if there's a better way to write it.Alternatively, note that 21√2/4 -1= (21√2 -4)/4. This is the exact form. But maybe the problem expects the answer to be written as 3√2 + (9√2)/4 -1= (12√2 +9√2)/4 -1=21√2/4 -1. Alternatively, factor out √2:√2*(21/4) -1= (21/4)√2 -1. But perhaps the answer is simply written as 21√2/4 -1.However, let me verify this once again.At x where cos x=1/3 and sin x=2√2/3:A(x)=√(10 -6*(1/3))=√(10-2)=√8=2√2B(x)=√(17/8 - (3√2)/2*(2√2/3))=√(17/8 - (3√2*2√2)/(6))=√(17/8 - (12/6))=√(17/8 -2)=√(17/8 -16/8)=√(1/8)=1/(2√2)=√2/4C(x)=√(19 -6√2)=3√2 -1Thus, f(x)=2√2 + √2/4 +3√2 -1= (2√2 +3√2) +√2/4 -1=5√2 +√2/4 -1= (20√2/4 +√2/4) -1=21√2/4 -1Yes, exactly. So the minimum value is 21√2/4 -1. To confirm that this is indeed the minimum, we can check the derivative at this point. However, since we derived this value by minimizing each term's contribution under the condition that they are achieved at the same x, and our numerical checks support this value, it's likely correct.Therefore, the minimal value is (21√2)/4 -1. To present this in the boxed form, we can write it as boxed{dfrac{21sqrt{2}}{4} - 1}.But let me check if this can be simplified further. For instance, 21/4 is 5.25, but since it's already in terms of √2, perhaps this is the simplest form.Alternatively, another approach: Let's consider that the minimal value occurs when the three terms correspond to the distances between points on the circles and their respective fixed points, aligned in a straight line. But this requires more advanced geometric analysis.Alternatively, using the geometric interpretation we had before:If we interpret each term as distances from points on three different circles (radii 3, 3/(2√2), and 3√2) to their respective fixed points (1,0), (0,1), and (1/3, 2√2/3), then the minimum total distance might be achieved when the three moving points are collinear with their fixed points in a way that the path through them is straight. This is similar to reflecting points in mirrors to find the shortest path, a common technique in optimization problems involving distances.However, with three points, the reflection method becomes complex. But perhaps considering two points first. For example, reflect one of the fixed points over the circle, find the distance, but this is non-trivial for multiple circles.Alternatively, since all three circles are parameterized by the same angle x, which implies that the three moving points are scaled versions of a single point on the unit circle. That is, if we let (cos x, sin x) be a point on the unit circle, then the three moving points are scaled by 3, 3/(2√2), and 3√2 respectively. So, they are scaled differently along the same direction x. Therefore, the problem reduces to finding the angle x such that the sum of the distances from these scaled points to their respective fixed points is minimized.This seems like a problem that could be approached using calculus, but the derivative is complicated. However, we found that at x where cos x=1/3 and sin x=2√2/3, the third term reaches its minimum, and at this angle, the sum of the terms gives a value of 21√2/4 -1, which seems to be the minimal value based on our calculations.Thus, after careful consideration and checking multiple angles, it appears that the minimal value of f(x) is indeed (21√2)/4 -1.Final AnswerThe minimum value of ( f(x) ) is (boxed{dfrac{21sqrt{2}}{4} - 1}).</think>To find the minimum value of the function ( f(x) ), we start by interpreting each term under the square roots as distances from points on circles to fixed points. The function is given by:[f(x) = sqrt{10 - 6 cos x} + sqrt{frac{17}{8} - frac{3 sqrt{2}}{2} sin x} + sqrt{19 - 2 sqrt{2} cos x - 8 sin x}]1. First Term: (sqrt{10 - 6 cos x}) represents the distance from the point ((3 cos x, 3 sin x)) to the fixed point ((1, 0)).2. Second Term: (sqrt{frac{17}{8} - frac{3 sqrt{2}}{2} sin x}) represents the distance from the point (left(frac{3}{2sqrt{2}} cos x, frac{3}{2sqrt{2}} sin xright)) to the fixed point ((0, 1)).3. Third Term: (sqrt{19 - 2 sqrt{2} cos x - 8 sin x}) represents the distance from the point ((3sqrt{2} cos x, 3sqrt{2} sin x)) to the fixed point (left(frac{1}{3}, frac{2sqrt{2}}{3}right)).We then evaluate the function at critical angles and use calculus to find the minimum value. By analyzing the terms and their geometric interpretations, we find that the minimum value occurs when the angle ( x ) satisfies ( cos x = frac{1}{3} ) and ( sin x = frac{2sqrt{2}}{3} ). This leads to the following calculations:- The first term simplifies to ( 2sqrt{2} ).- The second term simplifies to ( frac{sqrt{2}}{4} ).- The third term simplifies to ( 3sqrt{2} - 1 ).Summing these results, we get:[f(x) = 2sqrt{2} + frac{sqrt{2}}{4} + (3sqrt{2} - 1) = frac{21sqrt{2}}{4} - 1]Thus, the minimum value of ( f(x) ) is (boxed{dfrac{21sqrt{2}}{4} - 1}).,<think>Alright, so I need to find the minimum value of this function f(x). Let me write it down again to make sure I have it right:f(x) = sqrt(10 - 6 cos x) + sqrt(17/8 - (3√2)/2 sin x) + sqrt(19 - 2√2 cos x - 8 sin x)Hmm, okay. That's a sum of three square roots, each with different trigonometric expressions inside. Minimizing such a function seems tricky because each square root could have its own minimum at different x-values. But maybe there's a way to interpret these expressions geometrically or use some trigonometric identities to simplify them. Let me think.First, I recall that expressions under square roots that involve cos x and sin x can sometimes be rewritten using the identity A cos x + B sin x = C cos(x - φ) or something similar. Maybe that can help simplify each term. Let me try that for each term individually.Starting with the first term: sqrt(10 - 6 cos x). Let's see, 10 - 6 cos x. If I think of this as a constant minus a cosine term, maybe it's similar to the form sqrt(A - B cos x), which can be interpreted as the distance from a point in polar coordinates or something. Wait, actually, if we consider that sqrt((a cos x - b)^2 + (a sin x)^2), that's the distance from the point (b, 0) to the point (a cos x, a sin x) on a circle of radius a. So maybe each square root term represents a distance?Let me check. For example, take the first term: sqrt(10 - 6 cos x). Let's square it: 10 - 6 cos x. If this is the square of the distance between two points, then perhaps expanding the distance formula would give us this expression.Suppose we have two points: one at (sqrt(10), 0) and another at (3 cos x, 3 sin x). Then the distance squared between them would be (sqrt(10) - 3 cos x)^2 + (0 - 3 sin x)^2 = 10 - 6 sqrt(10) cos x + 9 cos²x + 9 sin²x. But that simplifies to 10 - 6 sqrt(10) cos x + 9(cos²x + sin²x) = 10 - 6 sqrt(10) cos x + 9 = 19 - 6 sqrt(10) cos x. Hmm, that's not the same as 10 - 6 cos x. Maybe I need different coordinates.Wait, maybe if the points are (something else). Let me try again. Suppose we have two points, (a, 0) and (b cos x, b sin x). The distance squared between them is (a - b cos x)^2 + (0 - b sin x)^2 = a² - 2ab cos x + b² cos²x + b² sin²x = a² - 2ab cos x + b²(cos²x + sin²x) = a² - 2ab cos x + b². So if I set a² + b² = 10 and 2ab = 6. Then, solving these equations:From 2ab = 6, we get ab = 3. Then, a² + b² = 10. Let me solve for a and b. Let's let a and b be positive real numbers. Then, (a + b)^2 = a² + 2ab + b² = 10 + 6 = 16, so a + b = 4. Similarly, (a - b)^2 = a² - 2ab + b² = 10 - 6 = 4, so a - b = 2. Then, solving a + b = 4 and a - b = 2 gives a = 3 and b = 1. So, a = 3, b = 1. Therefore, the distance between (3, 0) and (cos x, sin x) is sqrt((3 - cos x)^2 + (0 - sin x)^2) = sqrt(9 - 6 cos x + cos²x + sin²x) = sqrt(10 - 6 cos x). Perfect! So the first term is the distance between the points (3, 0) and (cos x, sin x).Cool, so the first term is the distance from (3, 0) to a point on the unit circle (cos x, sin x). Similarly, maybe the other terms can be interpreted as distances as well. Let me check the second term: sqrt(17/8 - (3√2)/2 sin x). Let's square it: 17/8 - (3√2)/2 sin x. Hmm, so similar approach: suppose this is the square of the distance between two points. Let's denote points (c, d) and (e cos x, e sin x). Then, distance squared is (c - e cos x)^2 + (d - e sin x)^2 = c² - 2ce cos x + e² cos²x + d² - 2de sin x + e² sin²x = c² + d² - 2ce cos x - 2de sin x + e²(cos²x + sin²x) = c² + d² + e² - 2ce cos x - 2de sin x. So we need to set this equal to 17/8 - (3√2)/2 sin x. Therefore, matching coefficients:-2ce = 0 (since there's no cos x term in the given expression)-2de = -3√2 / 2and c² + d² + e² = 17/8From -2ce = 0, we have either c = 0 or e = 0. But e can't be 0 because the point would collapse to the origin. So c = 0. Then, from -2de = -3√2 / 2, we have 2de = 3√2 / 2 => de = 3√2 / 4. Let e be the radius of the circle. Let's assume the point is on the unit circle again, so e = 1. Wait, but if e is 1, then d = 3√2 / 4. Then, c = 0, d = 3√2 / 4, e = 1. Then, c² + d² + e² = 0 + (9*2)/16 + 1 = 18/16 + 16/16 = 34/16 = 17/8. Perfect! So that works. Therefore, the second term is the distance between the point (0, 3√2 / 4) and the point (cos x, sin x) on the unit circle. Wait, but the radius here is 1? But the point (0, 3√2 / 4) is fixed. Wait, but in the distance formula, the moving point is on the unit circle. So the second term is the distance between (0, 3√2 /4) and (cos x, sin x). So yes, the second term is that distance.Similarly, let's check the third term: sqrt(19 - 2√2 cos x - 8 sin x). Squaring this gives 19 - 2√2 cos x - 8 sin x. Again, let's model this as the distance squared between a fixed point and a point on the unit circle. Let’s denote the fixed point as (p, q). Then, the distance squared is (p - cos x)^2 + (q - sin x)^2 = p² - 2p cos x + cos²x + q² - 2q sin x + sin²x = p² + q² - 2p cos x - 2q sin x + 1. Therefore, we need:p² + q² + 1 = 19 => p² + q² = 18-2p = -2√2 => p = √2-2q = -8 => q = 4So check if p = √2 and q = 4 satisfy p² + q² = (√2)^2 + 4^2 = 2 + 16 = 18. Yes! Perfect. Therefore, the third term is the distance between the point (√2, 4) and the point (cos x, sin x) on the unit circle.Therefore, putting this all together, the function f(x) is the sum of three distances:1. From (3, 0) to (cos x, sin x)2. From (0, 3√2 /4 ) to (cos x, sin x)3. From (√2, 4) to (cos x, sin x)So, geometrically, f(x) is the sum of the distances from three fixed points to a variable point on the unit circle. Therefore, to minimize f(x), we need to find the point P = (cos x, sin x) on the unit circle such that the sum of the distances from P to (3, 0), (0, 3√2 /4 ), and (√2, 4) is minimized.This is a classic optimization problem: minimizing the sum of distances from a moving point on a curve to several fixed points. For a single point, the minimum would just be the closest approach, but with multiple points, it's more complicated. This is similar to the Fermat-Torricelli problem, which seeks a point minimizing the sum of distances to three given points. However, in our case, the point P is constrained to lie on the unit circle.So, perhaps the approach is to consider reflecting some of the points across the unit circle or using Lagrange multipliers to find critical points. Alternatively, maybe there is a clever geometric interpretation where the three points and the unit circle have some symmetry that can be exploited.Alternatively, since all three distances are from points outside the unit circle (since (3,0) is at distance 3 from the origin, (0, 3√2 /4 ) is at distance 3√2 /4 ≈ 1.06, which is just outside the unit circle (since 3√2 /4 ≈ 1.06066), and (√2, 4) is at distance sqrt( (√2)^2 + 4^2 ) = sqrt(2 + 16) = sqrt(18) ≈ 4.24, so definitely outside). Therefore, all three fixed points are outside the unit circle.Hmm. So we need to find a point on the unit circle such that the sum of the distances to these three external points is minimized. That's non-trivial. Maybe using reflection? For instance, in problems where you have to find minimal paths reflecting off a surface, you reflect points across the surface. But here, since we have three points, reflecting all of them might not lead to a straightforward solution.Alternatively, perhaps we can parameterize the point on the unit circle as (cos x, sin x) and then write the total distance as a function of x, then take the derivative and set it to zero. But given that the function f(x) is already given in terms of x, maybe calculus is the way to go here. Let me try that.First, let's denote the three terms as follows:A(x) = sqrt(10 - 6 cos x)B(x) = sqrt(17/8 - (3√2)/2 sin x)C(x) = sqrt(19 - 2√2 cos x - 8 sin x)Then f(x) = A(x) + B(x) + C(x)To find the minimum, we can take the derivative f'(x) and set it equal to zero. Let's compute f'(x):f'(x) = A'(x) + B'(x) + C'(x)Compute each derivative:First, A(x) = sqrt(10 - 6 cos x)A'(x) = (1/(2 sqrt(10 - 6 cos x))) * (6 sin x) = (3 sin x) / sqrt(10 - 6 cos x)Second, B(x) = sqrt(17/8 - (3√2)/2 sin x)B'(x) = (1/(2 sqrt(17/8 - (3√2)/2 sin x))) * (- (3√2)/2 cos x) = (-3√2 cos x)/(4 sqrt(17/8 - (3√2)/2 sin x))Third, C(x) = sqrt(19 - 2√2 cos x - 8 sin x)C'(x) = (1/(2 sqrt(19 - 2√2 cos x - 8 sin x))) * (2√2 sin x - 8 cos x) = (2√2 sin x - 8 cos x)/(2 sqrt(19 - 2√2 cos x - 8 sin x)) = (√2 sin x - 4 cos x)/sqrt(19 - 2√2 cos x - 8 sin x)Therefore, setting f'(x) = 0:(3 sin x)/sqrt(10 - 6 cos x) - (3√2 cos x)/(4 sqrt(17/8 - (3√2)/2 sin x)) + (√2 sin x - 4 cos x)/sqrt(19 - 2√2 cos x - 8 sin x) = 0This equation looks quite complicated. Solving this analytically might be very difficult. Maybe we can look for critical points by testing specific angles x where the trigonometric functions take simple values, like x = 0, π/2, π/4, etc. Let me try plugging in some values.First, try x = 0:Compute A'(0): sin 0 = 0, so A'(0) = 0B'(0): cos 0 = 1, so numerator is -3√2 * 1 = -3√2. Denominator: 4 sqrt(17/8 - 0) = 4 sqrt(17/8) ≈ 4 * 1.456 ≈ 5.823. So B'(0) ≈ -3√2 / 5.823 ≈ -4.242 / 5.823 ≈ -0.728C'(0): sin 0 = 0, cos 0 = 1. Numerator: √2 * 0 - 4 * 1 = -4. Denominator: sqrt(19 - 2√2 *1 - 0) = sqrt(19 - 2.828) ≈ sqrt(16.172) ≈ 4.021. So C'(0) ≈ -4 / 4.021 ≈ -0.995Therefore, f'(0) ≈ 0 - 0.728 - 0.995 ≈ -1.723 < 0. So the derivative is negative at x=0. Therefore, the function is decreasing at x=0.Now try x = π/2:A'(π/2): sin(π/2) = 1. So numerator: 3*1=3. Denominator: sqrt(10 - 6*0)=sqrt(10)≈3.162. So A'(π/2)≈3 / 3.162≈0.949B'(π/2): cos(π/2)=0. So numerator: -3√2 * 0 = 0. So B'(π/2)=0C'(π/2): sin(π/2)=1, cos(π/2)=0. Numerator: √2*1 - 4*0 = √2 ≈1.414. Denominator: sqrt(19 -0 -8*1)=sqrt(11)≈3.316. So C'(π/2)≈1.414 /3.316≈0.426Therefore, f'(π/2)≈0.949 + 0 +0.426≈1.375 >0. So the derivative is positive at x=π/2. Therefore, the function is increasing at x=π/2.So between x=0 and x=π/2, the derivative goes from negative to positive, implying a minimum somewhere in between. Let's check x=π/4.Compute x=π/4:First, sin(π/4)=√2/2≈0.707, cos(π/4)=√2/2≈0.707.Compute A'(π/4):Numerator:3*(√2/2)≈2.121Denominator: sqrt(10 -6*(√2/2))=sqrt(10 -3√2)≈sqrt(10 -4.242)≈sqrt(5.758)≈2.398So A'(π/4)=2.121/2.398≈0.884B'(π/4):Numerator:-3√2*(√2/2)= -3*2/2= -3Denominator:4*sqrt(17/8 - (3√2)/2*(√2/2))=4*sqrt(17/8 - (3*2)/4)=4*sqrt(17/8 - 3/2)=4*sqrt(17/8 -12/8)=4*sqrt(5/8)=4*(√10)/4≈√10≈3.162Wait, let me compute step by step:Inside the square root:17/8 - (3√2)/2*(√2/2)First, (3√2)/2*(√2/2)=3*(2)/4=3/2=1.5So 17/8 - 1.5=17/8 -12/8=5/8≈0.625Thus, sqrt(5/8)=√(5)/√8=√10/(2*2)=√10/ (2*1.414)≈3.162/2.828≈1.118But denominator is 4*sqrt(5/8)=4*(√10)/ (2*2.828)≈4*(3.162)/5.656≈12.648/5.656≈2.236Wait, perhaps better exact calculation:Denominator:4*sqrt(5/8)=4*(√(5/8))=4*(√10)/ (2*2)=√10. Wait:sqrt(5/8)=sqrt(10)/ (2*sqrt(2))= (sqrt(10)/2)/sqrt(2)= sqrt(10)/ (2*sqrt(2))= sqrt(5)/ (2*1)= sqrt(5)/2? Wait:Wait, sqrt(5/8)=sqrt(10)/ (2*2)= sqrt(10)/4. No:Wait, 5/8=10/16, so sqrt(10/16)=sqrt(10)/4. So sqrt(5/8)=sqrt(10)/ (2*sqrt(2))= sqrt(5)/ (2*sqrt(2)/sqrt(2))= sqrt(5)/ (2*sqrt(2)/sqrt(2))... Wait, perhaps I messed up.Wait, 5/8 is 5 divided by 8. So sqrt(5/8)=sqrt(5)/sqrt(8)=sqrt(5)/(2*sqrt(2))= sqrt(5)/2.828≈2.236/2.828≈0.7906So denominator is 4*0.7906≈3.162Therefore, B'(π/4)= -3 /3.162≈-0.949C'(π/4):Numerator: √2*(√2/2) -4*(√2/2)= (2/2) - (4√2)/2=1 - 2√2≈1 -2.828≈-1.828Denominator: sqrt(19 -2√2*(√2/2) -8*(√2/2))=sqrt(19 -2*2/2 -8*(√2)/2)=sqrt(19 -2 -4√2)=sqrt(17 -4√2)≈sqrt(17 -5.656)≈sqrt(11.344)≈3.368Therefore, C'(π/4)= (-1.828)/3.368≈-0.543Therefore, total f'(π/4)=0.884 -0.949 -0.543≈0.884 -1.492≈-0.608 <0So at x=π/4, the derivative is still negative. Therefore, the function is decreasing here. So between x=π/4 and x=π/2, derivative goes from -0.608 to +1.375. Therefore, there's a critical point somewhere in between. Let's try x=π/3 (60 degrees).x=π/3≈1.047 radianssin(π/3)=√3/2≈0.866, cos(π/3)=0.5Compute A'(π/3):Numerator:3*(√3/2)≈2.598Denominator:sqrt(10 -6*0.5)=sqrt(10 -3)=sqrt(7)≈2.6458A'(π/3)=2.598 /2.6458≈0.981B'(π/3):Numerator:-3√2 *0.5≈-3*1.414*0.5≈-2.121Denominator:4*sqrt(17/8 - (3√2)/2*(√3/2))First, compute inside sqrt:17/8 - (3√2)/2*(√3/2)=17/8 - (3√6)/4≈2.125 - (3*2.449)/4≈2.125 -7.347/4≈2.125 -1.837≈0.288sqrt(0.288)=≈0.536Denominator:4*0.536≈2.144Thus, B'(π/3)= -2.121 /2.144≈-0.989C'(π/3):Numerator:√2*(√3/2) -4*0.5= (√6)/2 -2≈1.225 -2≈-0.775Denominator:sqrt(19 -2√2*0.5 -8*(√3/2))=sqrt(19 -√2 -4√3)Compute inside sqrt:√2≈1.414, 4√3≈6.928So 19 -1.414 -6.928≈19 -8.342≈10.658sqrt(10.658)≈3.266Thus, C'(π/3)= -0.775 /3.266≈-0.237Therefore, f'(π/3)=0.981 -0.989 -0.237≈0.981 -1.226≈-0.245 <0Still negative. So derivative at π/3 is -0.245. Let's try x=π/2. Let's check again.Wait, we already checked x=π/2, derivative was positive. Wait, but between π/3 and π/2, derivative goes from -0.245 to +1.375, so crosses zero. Let me pick x= 2π/5≈1.256 radians (72 degrees). Let's compute:x=2π/5≈1.256sin(2π/5)≈0.951, cos(2π/5)≈0.309Compute A'(2π/5):Numerator:3*0.951≈2.853Denominator:sqrt(10 -6*0.309)=sqrt(10 -1.854)=sqrt(8.146)≈2.854Thus, A'(2π/5)≈2.853 /2.854≈0.999≈1.0B'(2π/5):Numerator:-3√2 *0.309≈-3*1.414*0.309≈-1.312Denominator:4*sqrt(17/8 - (3√2)/2 *0.951)First, compute inside sqrt:17/8≈2.125(3√2)/2 *0.951≈3*1.414/2*0.951≈2.121*0.951≈2.017Thus, 2.125 -2.017≈0.108sqrt(0.108)≈0.329Denominator:4*0.329≈1.315Therefore, B'(2π/5)= -1.312 /1.315≈-0.998C'(2π/5):Numerator:√2 *0.951 -4*0.309≈1.345 -1.236≈0.109Denominator:sqrt(19 -2√2*0.309 -8*0.951)=sqrt(19 -0.876 -7.608)=sqrt(19 -8.484)=sqrt(10.516)≈3.243Thus, C'(2π/5)=0.109 /3.243≈0.0336Therefore, total f'(2π/5)=1.0 -0.998 +0.0336≈1.0 -0.998 +0.0336≈0.0356 >0So now, at x=2π/5, derivative is approximately +0.0356. So between x=π/3≈1.047 (where f'≈-0.245) and x=2π/5≈1.256 (f'≈+0.0356), the derivative crosses zero. Let me try x=1.2 radians (≈68.8 degrees):x=1.2sin(1.2)≈0.932, cos(1.2)≈0.362Compute A'(1.2):Numerator:3*0.932≈2.796Denominator:sqrt(10 -6*0.362)=sqrt(10 -2.172)=sqrt(7.828)≈2.798A'(1.2)=2.796 /2.798≈0.999≈1.0B'(1.2):Numerator:-3√2 *0.362≈-3*1.414*0.362≈-1.535Denominator:4*sqrt(17/8 - (3√2)/2 *0.932)First, inside sqrt:17/8≈2.125(3√2)/2 *0.932≈3*1.414/2 *0.932≈2.121*0.932≈1.976Thus, 2.125 -1.976≈0.149sqrt(0.149)≈0.386Denominator:4*0.386≈1.544Thus, B'(1.2)= -1.535 /1.544≈-0.994C'(1.2):Numerator:√2 *0.932 -4*0.362≈1.319 -1.448≈-0.129Denominator:sqrt(19 -2√2*0.362 -8*0.932)=sqrt(19 -1.025 -7.456)=sqrt(19 -8.481)=sqrt(10.519)≈3.243Thus, C'(1.2)= -0.129 /3.243≈-0.0398Total f'(1.2)=1.0 -0.994 -0.0398≈1.0 -1.0338≈-0.0338 <0So at x=1.2, f'≈-0.0338At x=2π/5≈1.256, f'≈+0.0356So the root is between 1.2 and 1.256. Let's use linear approximation.Between x=1.2 (f'=-0.0338) and x=1.256 (f'=+0.0356). The difference in x is 0.056, and the difference in f' is 0.0356 - (-0.0338)=0.0694. To find x where f'=0:x=1.2 + (0 - (-0.0338))/0.0694 *0.056≈1.2 + (0.0338 /0.0694)*0.056≈1.2 +0.487*0.056≈1.2 +0.027≈1.227So let's try x≈1.227 radians (≈70.3 degrees)Compute x=1.227sin(1.227)≈sin(70.3°)≈0.943, cos(1.227)≈cos(70.3°)≈0.333Compute A'(1.227):Numerator≈3*0.943≈2.829Denominator≈sqrt(10 -6*0.333)=sqrt(10 -2)=sqrt(8)≈2.828Thus, A'(1.227)=2.829 /2.828≈1.000B'(1.227):Numerator≈-3√2 *0.333≈-3*1.414*0.333≈-1.414Denominator:4*sqrt(17/8 - (3√2)/2 *0.943)Compute inside sqrt:(3√2)/2 *0.943≈3*1.414/2 *0.943≈2.121*0.943≈2.017/8≈2.125, so 2.125 -2.0≈0.125sqrt(0.125)=≈0.3536Denominator≈4*0.3536≈1.414Thus, B'(1.227)= -1.414 /1.414≈-1.0C'(1.227):Numerator≈√2*0.943 -4*0.333≈1.333 -1.332≈0.001Denominator≈sqrt(19 -2√2*0.333 -8*0.943)=sqrt(19 -0.942 -7.544)=sqrt(19 -8.486)=sqrt(10.514)≈3.243Thus, C'(1.227)=0.001 /3.243≈0.0003Therefore, total f'(1.227)=1.0 -1.0 +0.0003≈0.0003≈0. Close enough. Therefore, the critical point is near x≈1.227 radians.Therefore, the minimum occurs near x≈1.227. Let's compute f(x) at this x to approximate the minimal value.But this is a lot of work. Alternatively, maybe there's a geometric interpretation. Let me recall that the original function is the sum of three distances from points (3,0), (0, 3√2 /4 ), and (√2, 4) to a point on the unit circle. If we can find a point P on the unit circle such that P minimizes the sum of distances to these three points, then that's our answer.Alternatively, maybe use the method of reflection. For example, in the Fermat-Torricelli problem, the minimal sum of distances is achieved by a point such that the angles between the lines to the three points are all 120 degrees. However, in our case, the point is constrained to lie on the unit circle. So the reflection method may not apply directly.Alternatively, since all three fixed points are outside the unit circle, perhaps the minimal sum occurs at a point where the angles from P to each fixed point make equal angles with the gradient vectors or something like that. But this is vague.Alternatively, since we have three points, maybe the minimal sum occurs at the projection of the centroid of the three points onto the unit circle. But centroid is ( (3 +0 +√2)/3 , (0 + 3√2/4 +4)/3 ). Let's compute that:Centroid x-coordinate: (3 + √2)/3 ≈ (3 +1.414)/3≈4.414/3≈1.471Centroid y-coordinate: (0 +3√2/4 +4)/3≈(1.06 +4)/3≈5.06/3≈1.687Then, the centroid is at (≈1.471, ≈1.687). The distance from the origin to the centroid is sqrt(1.471² +1.687²)≈sqrt(2.163 +2.846)=sqrt(5.009)≈2.238. So the projection onto the unit circle would be (1.471/2.238, 1.687/2.238)≈(0.657, 0.754). Let's check what angle this corresponds to. The point (0.657,0.754) is in the first quadrant, angle≈arctan(0.754/0.657)≈arctan(1.147)≈48.9 degrees. But our critical point was at ≈70.3 degrees, so that's not matching. Maybe this approach is not valid.Alternatively, perhaps the minimal sum occurs where the point P is such that the sum of the unit vectors in the directions of the three fixed points is zero. But that might not necessarily hold.Alternatively, use complex numbers. Let me denote the point on the unit circle as e^{ix}. Then the three distances are |3 - e^{ix}|, | (3√2 /4)i - e^{ix}|, and |√2 +4i - e^{ix}|. But how does that help?Alternatively, use vector calculus. The gradient of the sum of distances should be tangent to the unit circle. But this is getting complicated.Alternatively, parametrize the point on the unit circle as (cos x, sin x) and plug back into the original function and compute the value at x≈1.227. Let's do that.Compute f(1.227):First term: sqrt(10 -6 cos x). cos x≈0.333, so 10 -6*0.333≈10 -2≈8. So sqrt(8)=2√2≈2.828Second term: sqrt(17/8 - (3√2)/2 sin x). sin x≈0.943. Compute (3√2)/2 *0.943≈2.121 *0.943≈2.0. So 17/8 -2≈2.125 -2=0.125. sqrt(0.125)=√(1/8)=√2/4≈0.3536Third term: sqrt(19 -2√2 cos x -8 sin x). cos x≈0.333, sin x≈0.943. Compute 2√2*0.333≈0.942, 8*0.943≈7.544. So 19 -0.942 -7.544≈19 -8.486≈10.514. sqrt(10.514)≈3.242Therefore, total f(x)=2.828 +0.3536 +3.242≈6.4236But is this the minimal value? Let's check the value at x≈1.227 gives f≈6.424. Let me check at x=π/4, which was another critical point candidate. Wait, no, x=π/4 was a point where the derivative was still negative.Wait, x=π/4≈0.785, but we found that the minimum is near x≈1.227. Let's check another point where maybe the three terms are minimized. Alternatively, check the case where the point P is aligned with one of the fixed points.For example, if P is (1,0), which is x=0. Then compute f(0):First term: sqrt(10 -6*1)=sqrt(4)=2Second term: sqrt(17/8 -0)=sqrt(17/8)=sqrt(2.125)=≈1.456Third term: sqrt(19 -2√2*1 -0)=sqrt(19 -2.828)=sqrt(16.172)=≈4.021Total f(0)=2 +1.456 +4.021≈7.477Compare to the critical point value≈6.424. So lower.Another point: x=π/2, P=(0,1). Compute f(π/2):First term: sqrt(10 -0)=sqrt(10)≈3.162Second term: sqrt(17/8 - (3√2)/2 *1)=sqrt(2.125 -2.121)=sqrt(0.004)=≈0.063Third term: sqrt(19 -0 -8*1)=sqrt(11)=≈3.316Total f(π/2)=3.162 +0.063 +3.316≈6.541This is ≈6.541, which is higher than 6.424. So the critical point near x≈1.227 gives a lower value. How about x=3π/4≈2.356?sin(3π/4)=√2/2≈0.707, cos(3π/4)=-√2/2≈-0.707First term: sqrt(10 -6*(-0.707))=sqrt(10 +4.242)=sqrt(14.242)=≈3.774Second term: sqrt(17/8 - (3√2)/2*(√2/2))=sqrt(17/8 -3/2)=sqrt(17/8 -12/8)=sqrt(5/8)=≈0.7906Third term: sqrt(19 -2√2*(-√2/2) -8*(√2/2))=sqrt(19 +2 -4√2)=sqrt(21 -5.656)=sqrt(15.344)=≈3.917Total f(3π/4)=3.774 +0.7906 +3.917≈8.481Higher again. How about x= arccos(1/3)≈1.230 radians (since in the critical point near x≈1.227, the cos x≈0.333≈1/3). Let's try x=arccos(1/3). Then cos x=1/3, sin x=sqrt(1 -1/9)=sqrt(8/9)=2√2/3≈0.9428Compute f(x):First term: sqrt(10 -6*(1/3))=sqrt(10 -2)=sqrt(8)=2√2≈2.828Second term: sqrt(17/8 - (3√2)/2*(2√2/3))=sqrt(17/8 - (3√2 * 2√2)/6)=sqrt(17/8 - (12)/6)=sqrt(17/8 -2)=sqrt(17/8 -16/8)=sqrt(1/8)=1/(2√2)≈0.3536Third term: sqrt(19 -2√2*(1/3) -8*(2√2/3))=sqrt(19 - (2√2 +16√2)/3)=sqrt(19 -18√2/3)=sqrt(19 -6√2)≈sqrt(19 -8.485)≈sqrt(10.515)≈3.242Therefore, total f(x)=2.828 +0.3536 +3.242≈6.424Which matches our previous approximation. So when x=arccos(1/3), which is approximately 1.230 radians, the function f(x)≈6.424. Moreover, the derivative at this point is approximately zero, so this is a critical point. Therefore, this is likely the minimum.Wait, but how did we get exact values here? When I took x=arccos(1/3), then sin x=2√2/3. Let's verify:First term: sqrt(10 -6*(1/3))=sqrt(10 -2)=sqrt(8)=2√2Second term: sqrt(17/8 - (3√2)/2*(2√2/3)). Let's compute the expression inside:(3√2)/2 * (2√2)/3 = (3*2*√2*√2)/(2*3) = (6*2)/6=2. So 17/8 -2=17/8 -16/8=1/8. Thus, sqrt(1/8)=1/(2√2)=√2/4≈0.3536.Third term: sqrt(19 -2√2*(1/3) -8*(2√2/3))=sqrt(19 - (2√2 +16√2)/3)=sqrt(19 -18√2/3)=sqrt(19 -6√2). Compute 19 -6√2≈19 -8.485≈10.515, sqrt of that≈3.242Therefore, exact values:First term: 2√2Second term: √(1/8)=√2/4Third term: sqrt(19 -6√2)So total f(x)=2√2 +√2/4 +sqrt(19 -6√2)Is there a way to simplify sqrt(19 -6√2)? Let's check if 19 -6√2 is a square. Suppose sqrt(a - b√2)=sqrt(c) - sqrt(d). Let's square both sides:a - b√2 = c + d - 2 sqrt(c d)Comparing terms:c + d = a-2 sqrt(c d)= -b√2 => 2 sqrt(c d)=b√2 => sqrt(c d)=b√2 /2 => c d = (b² *2)/4 = b² /2So for our case, a=19, b=6. Therefore,c + d =19c d=6² /2=36/2=18So solving for c and d:We need two numbers c and d such that c + d =19 and c*d=18. The solutions would be roots of x² -19x +18=0. The discriminant is 361 -72=289, sqrt(289)=17. Thus, c=(19±17)/2. So c=(19+17)/2=36/2=18, or c=(19-17)/2=2/2=1. Therefore, c=18 and d=1, or vice versa.Therefore, sqrt(19 -6√2)=sqrt(18) - sqrt(1)=3√2 -1. Let me verify:(3√2 -1)^2= (3√2)^2 -2*3√2*1 +1^2=18 -6√2 +1=19 -6√2. Perfect!Therefore, sqrt(19 -6√2)=3√2 -1Therefore, f(x)=2√2 +√2/4 +3√2 -1= (2√2 +3√2 +√2/4) -1= (5√2 +√2/4) -1= (20√2/4 +√2/4) -1= (21√2)/4 -1Therefore, the minimal value of f(x) is (21√2)/4 -1Let me compute this numerically:21√2≈21*1.414≈29.69429.694 /4≈7.42357.4235 -1=6.4235, which matches our approximate calculation earlier.So, the minimum value is (21√2)/4 -1. To write this as a combined fraction:21√2 /4 -1= (21√2 -4)/4But let me check:(21√2 -4)/4=21√2/4 -4/4=21√2/4 -1. Yes.Alternatively, perhaps rationalizing or further simplification, but I think (21√2)/4 -1 is the simplest form. Let me verify the steps again to make sure.First, we found that at x=arccos(1/3), the three terms become:First term: 2√2Second term: √2/4Third term: 3√2 -1Therefore, adding them up:2√2 +√2/4 +3√2 -1= (2√2 +3√2) +√2/4 -1=5√2 +√2/4 -1= (5*4√2 +√2)/4 -1= (20√2 +√2)/4 -1=21√2/4 -1. Correct.Therefore, the minimal value is (21√2)/4 -1. Let me check if this can be written differently or simplified further. Since 21 and 4 are coprime, and √2 is irrational, this is as simplified as it gets.Therefore, the minimum value of f(x) is (21√2)/4 -1. To put it in the requested boxed format, we can write:boxed{dfrac{21sqrt{2}}{4} - 1}Final AnswerThe minimum value of ( f(x) ) is (boxed{dfrac{21sqrt{2}}{4} - 1}).</think>To find the minimum value of the function ( f(x) ), we start by interpreting each term geometrically as distances from a point on the unit circle to three fixed points. The function is given by:[f(x) = sqrt{10 - 6 cos x} + sqrt{frac{17}{8} - frac{3 sqrt{2}}{2} sin x} + sqrt{19 - 2 sqrt{2} cos x - 8 sin x}]We recognize each term as the distance from a point on the unit circle ((cos x, sin x)) to three fixed points:1. The first term is the distance from ((3, 0)) to ((cos x, sin x)).2. The second term is the distance from ((0, frac{3sqrt{2}}{4})) to ((cos x, sin x)).3. The third term is the distance from ((sqrt{2}, 4)) to ((cos x, sin x)).We then use calculus to find the critical points by setting the derivative ( f'(x) ) to zero. After evaluating the derivative at several points and using approximations, we find that the critical point near ( x approx 1.227 ) radians (which corresponds to ( arccos(1/3) )) gives the minimum value.At ( x = arccos(1/3) ), we compute the exact values of each term:1. The first term simplifies to ( 2sqrt{2} ).2. The second term simplifies to ( frac{sqrt{2}}{4} ).3. The third term simplifies to ( 3sqrt{2} - 1 ).Summing these values, we get:[2sqrt{2} + frac{sqrt{2}}{4} + 3sqrt{2} - 1 = frac{21sqrt{2}}{4} - 1]Thus, the minimum value of ( f(x) ) is:[boxed{dfrac{21sqrt{2}}{4} - 1}]]

🔑:8. frac{21 sqrt{2}}{4}-1.Noticebegin{aligned}f(x)= & sqrt{(cos x-3)^{2}+sin ^{2} x}+ & sqrt{cos ^{2} x+left(sin x-frac{3 sqrt{2}}{4}right)^{2}}+ & sqrt{(cos x-sqrt{2})^{2}+(sin x-4)^{2}} .end{aligned}Let A(3,0) 、 Bleft(0, frac{3 sqrt{2}}{4}right) 、 C(sqrt{2}, 4) 、 P(cos x, sin x). Then,f(x)=|P A|+|P B|+|P C|,where point P is on the circle odot O: x^{2}+y^{2}=1.Since the circle odot O: x^{2}+y^{2}=1 is tangent to A B at point Dleft(frac{1}{3}, frac{2 sqrt{2}}{3}right), and points O 、 D 、 C are collinear, we haveP C geqslant D C text {. }Also, P B+P A geqslant D B+D A=A B, sobegin{array}{l}P A+P B+P C geqslant A B+C D =frac{9 sqrt{2}}{4}+(3 sqrt{2}-1)=frac{21 sqrt{2}}{4}-1 .end{array}

🧠:[<think>Okay, so I need to solve these two problems. Let's start with the first one: comparing the sizes of y/x and (y + m)/(x + m) given that x > y > 0 and m > 0. Hmm, alright. Let me think. First, I remember that when you add the same positive number to both the numerator and denominator of a fraction, the fraction gets closer to 1. Since x > y, the original fraction y/x is less than 1. So adding m to both should make the new fraction larger than the original one. Let me verify that.Let's take an example. Suppose x = 5, y = 3, m = 2. Then y/x is 3/5 = 0.6, and (y + m)/(x + m) is (3 + 2)/(5 + 2) = 5/7 ≈ 0.714. So indeed, 5/7 is larger than 3/5. Another example: x = 10, y = 1, m = 1. Then y/x is 1/10 = 0.1, and (1 + 1)/(10 + 1) = 2/11 ≈ 0.1818, which is also larger. So my initial thought seems correct. But how do I prove it generally?Let's consider the difference between (y + m)/(x + m) and y/x. To compare them, subtract one from the other:(y + m)/(x + m) - y/x = [x(y + m) - y(x + m)] / [x(x + m)]Simplifying the numerator: xy + xm - xy - ym = xm - ym = m(x - y)Since x > y, m > 0, so the numerator is positive. The denominator is x(x + m), which is also positive because x and m are positive. Therefore, the entire expression is positive, which means (y + m)/(x + m) - y/x > 0. Therefore, (y + m)/(x + m) > y/x. So part 1 is done.Now part 2: Prove analytically that √(xy)(2 - √(xy)) ≤ 1. Hmm. Let me try to understand this inequality. Let's denote √(xy) as a variable, say t. Then the left-hand side becomes t(2 - t). So we have t(2 - t) ≤ 1. Let's analyze this quadratic function.t(2 - t) = -t² + 2t. So we need to show that -t² + 2t ≤ 1. Let's rearrange the inequality:-t² + 2t - 1 ≤ 0 → - (t² - 2t + 1) ≤ 0 → - (t - 1)² ≤ 0. Since (t - 1)² is always non-negative, multiplying by -1 makes it non-positive. Therefore, - (t - 1)² ≤ 0 is always true. Hence, the inequality holds for all real numbers t. Therefore, √(xy)(2 - √(xy)) ≤ 1 is always true, regardless of the values of x and y (as long as √(xy) is real, which requires x and y to be non-negative, which they are since x > y > 0). But the problem says "Prove analytically". Maybe they want a more detailed proof. Let me try expanding from the start.Starting with the left-hand side:√(xy)(2 - √(xy)) = 2√(xy) - (√(xy))² = 2√(xy) - xy.So we need to show that 2√(xy) - xy ≤ 1.Let me consider the function f(t) = 2t - t² where t = √(xy). Then f(t) = -t² + 2t. As before, completing the square: f(t) = - (t² - 2t) = - (t² - 2t + 1 - 1) = - (t - 1)^2 + 1. Therefore, f(t) = - (t - 1)^2 + 1. Since (t - 1)^2 is non-negative, the maximum value of f(t) is 1, achieved when t = 1. Therefore, f(t) ≤ 1 for all real t, which means 2√(xy) - xy ≤ 1. Therefore, the inequality holds. But maybe another approach using AM-GM inequality? Let's see. The original expression is 2√(xy) - xy. Let me set a = √(xy). Then the expression is 2a - a². To find its maximum, take derivative: d/da (2a - a²) = 2 - 2a. Setting derivative to zero: 2 - 2a = 0 → a = 1. Then maximum value is 2(1) - 1² = 1. Therefore, maximum value is 1, hence 2a - a² ≤ 1 for all real a. So the inequality holds. Alternatively, maybe using substitution. Let’s suppose that we can write the inequality as:√(xy)(2 - √(xy)) ≤ 1Let’s rearrange terms:√(xy) * 2 - (√(xy))² ≤ 1Which is:2√(xy) - xy ≤ 1Then, moving all terms to left:2√(xy) - xy - 1 ≤ 0Let’s denote t = √(xy). Then the inequality becomes:2t - t² - 1 ≤ 0 → -t² + 2t -1 ≤ 0Which factors as -(t² - 2t +1) = -(t -1)^2 ≤ 0. Which is true because squares are non-negative, so negative squares are non-positive. Hence, the inequality holds for all t. Therefore, the original inequality holds for all x, y > 0. Therefore, the analytic proof is complete. Let me check if I missed any constraints. Since x > y > 0, √(xy) is a real positive number. So t is positive, but the proof holds for all real t. So even if t is positive, the conclusion remains. Therefore, the inequality is always true. So summarizing:1. (y + m)/(x + m) is greater than y/x because the difference is positive.2. The expression √(xy)(2 - √(xy)) is always less than or equal to 1 because it can be rewritten as a negative square plus 1, which is always ≤ 1.Final Answer1. boxed{dfrac{y}{x} < dfrac{y + m}{x + m}}2. boxed{sqrt{xy}(2 - sqrt{xy}) leq 1}</think>Given ( x > y > 0 ) and ( m > 0 ):1. To compare the sizes of (frac{y}{x}) and (frac{y+m}{x+m}): Consider the difference between (frac{y+m}{x+m}) and (frac{y}{x}): [ frac{y+m}{x+m} - frac{y}{x} = frac{x(y+m) - y(x+m)}{x(x+m)} = frac{xy + xm - xy - ym}{x(x+m)} = frac{m(x - y)}{x(x+m)} ] Since ( x > y ), ( m > 0 ), and ( x ) and ( x + m ) are positive, the numerator and denominator are positive. Therefore, the difference is positive, implying: [ frac{y+m}{x+m} > frac{y}{x} ]2. To prove analytically that (sqrt{xy}(2 - sqrt{xy}) leq 1): Let ( t = sqrt{xy} ). The expression becomes: [ t(2 - t) = 2t - t^2 ] Rewrite the quadratic expression: [ 2t - t^2 = -t^2 + 2t = -(t^2 - 2t) = -(t^2 - 2t + 1 - 1) = -(t - 1)^2 + 1 ] Since ((t - 1)^2) is non-negative, (-(t - 1)^2) is non-positive. Therefore: [ -(t - 1)^2 + 1 leq 1 ] Hence, the inequality holds for all real ( t ), which means: [ sqrt{xy}(2 - sqrt{xy}) leq 1 ] Final Answer1. boxed{dfrac{y}{x} < dfrac{y + m}{x + m}}2. boxed{sqrt{xy}(2 - sqrt{xy}) leq 1},<think>Okay, so I have two math problems to solve here. Let me start with the first one. Problem 1: Compare the sizes of y/x and (y + m)/(x + m) where x > y > 0 and m > 0.Hmm, I need to figure out if y/x is less than, greater than, or equal to (y + m)/(x + m). Let me think. Since x and y are positive numbers with x bigger than y, and m is a positive number being added to both numerator and denominator. Wait, I remember something about fractions: if you add the same positive number to both the numerator and the denominator of a fraction less than 1, the fraction increases. Since x > y, then y/x is less than 1. Adding m to both y and x would make the new fraction (y + m)/(x + m). So according to that rule, this should be larger than y/x. Let me check with some numbers to be sure.Suppose x = 5, y = 3, m = 2. Then y/x = 3/5 = 0.6. (y + m)/(x + m) = 5/7 ≈ 0.714. So indeed, 0.714 > 0.6. Another example: x = 4, y = 1, m = 1. Then y/x = 1/4 = 0.25. (1 + 1)/(4 + 1) = 2/5 = 0.4. Still larger. What if m is really large? Say x = 5, y = 3, m = 100. Then (3 + 100)/(5 + 100) = 103/105 ≈ 0.98095. Original was 0.6, so still larger. Alternatively, maybe we can subtract the two fractions to see the difference. Let's compute (y + m)/(x + m) - y/x. First, common denominator is x(x + m). So:[(y + m)x - y(x + m)] / [x(x + m)] = [xy + mx - xy - ym] / [x(x + m)] = [m(x - y)] / [x(x + m)].Since m > 0, x > y (given), so numerator is positive. Denominator is positive because x and m are positive. So the entire expression is positive. Therefore, (y + m)/(x + m) - y/x > 0, which means (y + m)/(x + m) > y/x. So the answer for part 1 is that y/x is less than (y + m)/(x + m).Problem 2: Prove analytically that √(xy)(2 - √(xy)) ≤ 1.Alright, I need to prove that the expression √(xy)(2 - √(xy)) is less than or equal to 1. Let me first understand the expression. Let me denote √(xy) as a single variable to simplify. Let’s say t = √(xy). Then the expression becomes t(2 - t) = 2t - t². So the problem is to show that 2t - t² ≤ 1.But since t = √(xy), and given that x > y > 0, maybe there's some constraint on t? Wait, actually, x and y are positive numbers, but the problem statement for part 2 doesn't specify any constraints. Wait, is this problem 2 related to problem 1? The original given is x > y > 0, m > 0. Wait, but problem 2 is separate. Wait, let me check the original problem again.Wait, the user said: "Given x > y > 0, m > 0. 1. Compare... 2. Prove..." So both problems are under the same given conditions? Wait, but problem 2 is about √(xy)(2 - √(xy)) ≤ 1. Let me see.Wait, the original problem states "Given x > y > 0, m > 0." Then two separate problems. So for problem 2, is x and y the same as in problem 1? Probably yes. So we have x > y > 0, m > 0. But in problem 2, the inequality doesn't involve m. Hmm. So maybe there's a missing constraint? Or perhaps there's an assumption that x and y are such that √(xy) is in a certain range?Wait, let me think. The expression √(xy)(2 - √(xy)). Let me set t = √(xy) as before. Then the expression is 2t - t². The maximum value of 2t - t² is achieved when derivative is zero. Let's take derivative: d/dt (2t - t²) = 2 - 2t. Setting to zero: 2 - 2t = 0 => t = 1. Then maximum value is 2(1) - (1)^2 = 2 - 1 = 1. Therefore, 2t - t² ≤ 1 for all real t. Wait, is that true?Wait, because 2t - t² is a quadratic that opens downward, vertex at t=1, maximum value 1. So for any real number t, 2t - t² ≤ 1. Therefore, as long as t is a real number, the inequality holds. Since √(xy) is a real number (because x and y are positive), then the inequality √(xy)(2 - √(xy)) ≤ 1 holds for any positive x and y. So maybe the given conditions x > y > 0 are there for problem 1, but problem 2 is a general inequality that holds for any positive x, y?But why then is the problem phrased as "Given x > y > 0, m > 0"? Maybe problem 2 is under the same conditions? But m is not involved. Hmm.Wait, let me check the problem again. The user wrote:"Given x > y > 0, m > 0.1. Compare the sizes of y/x and (y + m)/(x + m).2. Prove analytically that √(xy)(2 - √(xy)) ≤ 1."So the same given conditions apply to both problems. So problem 2 must be under the same x > y > 0, m > 0. But the inequality in problem 2 doesn't involve m. So maybe there's a typo, or maybe the problem 2 is independent? Or maybe in problem 2, it's using some relationship from problem 1? Not sure.Alternatively, maybe the problem 2 is a separate problem, also under the same initial conditions. But since m isn't in the expression, perhaps we need to prove the inequality √(xy)(2 - √(xy)) ≤ 1 given that x > y > 0. But even so, the inequality seems general. Let me check with numbers.Suppose x = 4, y = 1. Then √(4*1) = 2. Then 2*(2 - 2) = 0 ≤ 1. Okay. If x = y, but x > y is given, so x can't equal y. Let me take x = 3, y = 1. Then √(3*1) = √3 ≈ 1.732. Then 1.732*(2 - 1.732) ≈ 1.732*(0.268) ≈ 0.464 ≤ 1. Another example: x = 2, y = 1. Then √2 ≈ 1.414. 1.414*(2 - 1.414) ≈ 1.414*0.586 ≈ 0.828 ≤ 1. If x approaches y, say x = y + ε where ε is very small. Then √(xy) = √(y(y + ε)) ≈ y + ε/2. Then the expression becomes (y + ε/2)(2 - y - ε/2). But unless y is constrained, how does that relate?Wait, but the given conditions are x > y > 0, but x and y can be any positive numbers as long as x is bigger. So maybe √(xy) can be any positive real number? Wait, but for x > y > 0, √(xy) is the geometric mean. However, x and y can be scaled. For example, if we set x = ty where t > 1, then √(xy) = √(ty * y) = y√t. So depending on y and t, √(xy) can be any positive number. Wait, but if x and y are arbitrary positive numbers with x > y, then √(xy) can be any positive number. For example, fix y = 1, then x can be any number greater than 1, so √(x*1) can be any number greater than 1. Wait, but in that case, if √(xy) is greater than 1, then 2 - √(xy) is less than 1, so √(xy)*(2 - √(xy)) would be less than √(xy)*1. But if √(xy) > 1, then this product is greater than 1. Wait, but that contradicts the inequality we need to prove. Wait, but in my previous example where x = 4, y = 1, √(xy) = 2, then 2*(2 - 2) = 0 ≤ 1. Wait, but if x = 9, y = 1, then √(9*1)=3. So 3*(2 - 3) = 3*(-1) = -3 ≤ 1. That also satisfies. Wait, but if x is such that √(xy) is greater than 2, then 2 - √(xy) is negative, so the product is negative, which is definitely less than 1. If √(xy) is between 0 and 2, then 2t - t² is a quadratic with maximum at t=1, so it reaches 1 there and is less otherwise. If t is greater than 2, then 2t - t² is negative. So in all cases, 2t - t² ≤ 1. Therefore, the inequality holds for any real t, but in our case t = √(xy) is positive. Therefore, the inequality holds for any positive x and y. So maybe the given conditions x > y > 0 are just part of the problem setup but not necessary for problem 2? Or maybe there's an error in the problem statement?Wait, but the problem says "prove analytically that √(xy)(2 - √(xy)) ≤ 1". Since we can show that 2t - t² ≤ 1 for all real t, which is equivalent to -t² + 2t -1 ≤ 0 => -(t² - 2t +1) ≤ 0 => -(t -1)^2 ≤ 0. Since squares are non-negative, -(t -1)^2 is non-positive, so the inequality holds for all real t. Therefore, √(xy)(2 - √(xy)) = -(√(xy) -1)^2 +0 ≤ 0 +1? Wait, wait. Let me rearrange the quadratic.Wait, 2t - t² = -t² + 2t = -(t² - 2t) = -(t² - 2t +1 -1) = -(t -1)^2 +1. So 2t - t² = 1 - (t -1)^2. Therefore, since (t -1)^2 ≥ 0, then 1 - (t -1)^2 ≤ 1. Therefore, 2t - t² ≤1 for all real t. Hence, √(xy)(2 - √(xy)) ≤1. Therefore, the inequality holds for any real number √(xy), but since x and y are positive, √(xy) is positive real, so the inequality holds. Therefore, the proof is straightforward by completing the square or recognizing the quadratic's maximum.So maybe the problem is just testing recognizing that it's a quadratic in terms of √(xy) and then finding its maximum. So the answer is that it's always less than or equal to 1, achieved when √(xy) =1, i.e., when xy=1. But given that x > y >0, maybe the equality holds when xy=1? Wait, if x > y and xy =1, then x > y and y =1/x. So x >1/x, which implies x² >1, so x >1. So in that case, there are x and y where equality holds.Therefore, the analytical proof would involve expressing the left-hand side as 2t - t² where t =√(xy), then showing that 2t - t² ≤1 for all real t, which can be done by rewriting it as 1 - (t -1)^2 ≤1, which is always true since (t -1)^2 ≥0.Alternatively, using calculus: take derivative of 2t - t², find maximum at t=1, value 1, hence 2t - t² ≤1.Alternatively, using AM-GM inequality. Wait, but I don't see a direct connection. Maybe not. Alternatively, note that 2t - t² -1 = - (t² - 2t +1) = -(t -1)^2 ≤0, so 2t -t² ≤1.Therefore, the proof is straightforward once you substitute t = √(xy). But the problem says "prove analytically", so maybe they want a more detailed step-by-step proof rather than just stating it's a quadratic. Let me structure it properly.Proof for Problem 2:Let t = √(xy). Since x and y are positive real numbers, t is also a positive real number.Consider the expression t(2 - t). Let's rewrite it:t(2 - t) = 2t - t².We need to show that 2t - t² ≤ 1 for all positive real numbers t.Rearrange the inequality:2t - t² ≤ 1 => -t² + 2t - 1 ≤ 0 => -(t² - 2t + 1) ≤ 0 => -(t - 1)² ≤ 0.Since (t - 1)² is always non-negative for real t, multiplying by -1 gives a non-positive result. Therefore, -(t - 1)² ≤ 0 holds for all real t. Hence, 2t - t² ≤ 1 for all real t. Substituting back t = √(xy), we have:√(xy)(2 - √(xy)) ≤ 1.This completes the proof. Equality holds when (t - 1)² = 0, i.e., when t = 1, which means √(xy) = 1, so xy = 1.Therefore, the inequality holds for all positive x and y, with equality if and only if xy = 1.Wait, but the original problem didn't specify that xy =1 is allowed. Given that x > y >0, can xy=1? For example, x=2, y=0.5. Then x=2 > y=0.5, and xy=1. So yes, equality is achievable under the given conditions. Therefore, the inequality is tight.So I think that's the solution.Final Answer1. boxed{dfrac{y}{x} < dfrac{y + m}{x + m}}2. boxed{sqrt{xy}(2 - sqrt{xy}) leq 1}</think>Given ( x > y > 0 ) and ( m > 0 ):1. Compare the sizes of (frac{y}{x}) and (frac{y+m}{x+m}): To determine the relationship between (frac{y}{x}) and (frac{y+m}{x+m}), we consider the difference between the two fractions: [ frac{y + m}{x + m} - frac{y}{x} = frac{(y + m)x - y(x + m)}{x(x + m)} = frac{m(x - y)}{x(x + m)} ] Since ( m > 0 ), ( x > y ), and the denominator ( x(x + m) > 0 ), the numerator ( m(x - y) ) is positive. Therefore, the difference is positive, implying: [ frac{y + m}{x + m} > frac{y}{x} ] Thus, the answer is (boxed{dfrac{y}{x} < dfrac{y + m}{x + m}}).2. Prove analytically that (sqrt{xy}(2 - sqrt{xy}) leq 1): Let ( t = sqrt{xy} ). The expression becomes: [ t(2 - t) = 2t - t^2 ] We need to show ( 2t - t^2 leq 1 ). Rearranging the inequality: [ 2t - t^2 leq 1 implies -t^2 + 2t - 1 leq 0 implies -(t - 1)^2 leq 0 ] Since ((t - 1)^2) is always non-negative, (-(t - 1)^2) is non-positive. Therefore, the inequality holds for all real ( t ). Substituting back ( t = sqrt{xy} ), we get: [ sqrt{xy}(2 - sqrt{xy}) leq 1 ] Equality holds when ( sqrt{xy} = 1 ), i.e., ( xy = 1 ). Thus, the inequality is proven analytically: [ boxed{sqrt{xy}(2 - sqrt{xy}) leq 1} ]]

🔑:1. We need to compare frac{y}{x} and frac{y+m}{x+m}. Let's subtract one from the other:begin{align*}frac{y}{x} - frac{y+m}{x+m} &= frac{yx + ym - xy - xm}{x(x+m)} &= frac{ym - xm}{x(x+m)} &= frac{m(y - x)}{x(x+m)}.end{align*}Since x > y > 0, we know that y - x 0 and x > 0, we have m(y-x) 0. Therefore, frac{m(y-x)}{x(x+m)} 0. Therefore, it is proven that sqrt{xy}(2 - sqrt{xy}) leq 1.boxed{sqrt{xy}(2 - sqrt{xy}) leq 1}.

🧠:[<think>Okay, so I need to solve this system of equations:[begin{cases}x^3 + y^3 = 35 x + y = 5end{cases}]Hmm, let's see. I remember there are formulas for factoring sums of cubes. Maybe that can help here. The sum of cubes formula is (a^3 + b^3 = (a + b)(a^2 - ab + b^2)). So maybe I can apply that to the first equation.Given that (x + y = 5), if I use the sum of cubes formula on (x^3 + y^3), it becomes:[(x + y)(x^2 - xy + y^2) = 35]Since (x + y = 5), substitute that in:[5(x^2 - xy + y^2) = 35]Divide both sides by 5:[x^2 - xy + y^2 = 7]Now, I have another equation from (x + y = 5). Maybe I can express one variable in terms of the other and substitute. Let's solve for (y) in terms of (x):[y = 5 - x]Then substitute this into the equation (x^2 - xy + y^2 = 7):First, let's compute each term:(y = 5 - x), so (y^2 = (5 - x)^2 = 25 - 10x + x^2)(xy = x(5 - x) = 5x - x^2)So substituting into (x^2 - xy + y^2):[x^2 - (5x - x^2) + (25 - 10x + x^2)]Let's simplify term by term:First term: (x^2)Second term: (-5x + x^2) because of the negative sign outside the parentheses.Third term: (25 - 10x + x^2)So combining all together:(x^2 - 5x + x^2 + 25 - 10x + x^2)Combine like terms:(x^2 + x^2 + x^2 = 3x^2)(-5x -10x = -15x)And the constant term is 25.So the equation becomes:[3x^2 -15x +25 =7]Subtract 7 from both sides:[3x^2 -15x +18 =0]Let me check if I did the algebra correctly. Let me retrace:Original substitution:(x^2 - (5x - x^2) + (25 -10x +x^2))So:(x^2 -5x + x^2 +25 -10x +x^2)Yes, that's 3x² -15x +25. Then set equal to 7, so 3x² -15x +18=0.Yes, that's correct.So, simplifying the quadratic equation:Divide all terms by 3:(x^2 -5x +6=0)Wait, 18 divided by 3 is 6, yes. So:(x^2 -5x +6=0)Factor this quadratic:Looking for two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3.So, ((x - 2)(x - 3) = 0)Therefore, solutions are (x = 2) and (x = 3).Since (x + y = 5), if (x = 2), then (y = 3). If (x = 3), then (y = 2).So the solutions are the pairs (2, 3) and (3, 2).Let me check these solutions in the original equation to make sure.First, check (2,3):(x^3 + y^3 = 8 + 27 = 35), which matches.(x + y = 5), which also matches.Then check (3,2):Same as above, (3^3 +2^3 =27 +8=35), and (3 +2=5). Both equations satisfied.Therefore, the solutions are (2,3) and (3,2).Wait, but the problem is presented as number 6.078, so maybe expecting just the answer? But let me confirm once more.Wait, maybe there is another way to approach this. Alternatively, using substitution from the start.Given (x + y =5), and (x^3 + y^3 =35). Let me recall that (x^3 + y^3 = (x + y)^3 - 3xy(x + y)). Maybe that identity is helpful here.Yes, indeed. Let's use that.So, (x^3 + y^3 = (x + y)^3 - 3xy(x + y)). Given that (x + y =5), substitute:(35 = 5^3 - 3xy(5))Compute (5^3 =125), so:(35 =125 -15xy)Then, subtract 125 from both sides:(35 -125 = -15xy)(-90 = -15xy)Divide both sides by -15:(6 = xy)So, (xy=6)Therefore, we have (x + y=5) and (xy=6). So these are the sum and product of two numbers. Therefore, the quadratic equation formed by these is:(t^2 -5t +6=0), same as before, leading to t=2 and t=3. Hence, same solutions.So, either way, the solutions are (2,3) and (3,2).Therefore, the answer should be these ordered pairs.But let me check if there are any other possible solutions. Since the system consists of a cubic and a linear equation, in general, there could be up to three solutions. But here, since we have factored the cubic into a quadratic and a linear equation, which then reduces to a quadratic. So, the quadratic can have at most two real solutions, which we found. Thus, these are the only solutions.Alternatively, maybe complex solutions, but since the problem is given without specifying the domain, likely expecting real solutions. Also, original equations have real coefficients, but even if complex, these would be the only solutions. But given that x + y =5, which is real, and x^3 + y^3=35, which is real, so x and y are likely real numbers.Hence, the solutions are (2,3) and (3,2).Final AnswerThe solutions are boxed{(2, 3)} and boxed{(3, 2)}.</think>Given the system of equations:[begin{cases}x^3 + y^3 = 35 x + y = 5end{cases}]We start by using the sum of cubes factorization on the first equation:[x^3 + y^3 = (x + y)(x^2 - xy + y^2)]Substituting (x + y = 5) into the equation, we get:[35 = 5(x^2 - xy + y^2)]Dividing both sides by 5:[7 = x^2 - xy + y^2]Next, we express (y) in terms of (x) from the second equation:[y = 5 - x]Substituting (y = 5 - x) into the equation (x^2 - xy + y^2 = 7):[x^2 - x(5 - x) + (5 - x)^2 = 7]Expanding and simplifying:[x^2 - 5x + x^2 + 25 - 10x + x^2 = 7]Combining like terms:[3x^2 - 15x + 25 = 7]Subtracting 7 from both sides:[3x^2 - 15x + 18 = 0]Dividing by 3:[x^2 - 5x + 6 = 0]Factoring the quadratic equation:[(x - 2)(x - 3) = 0]Thus, the solutions for (x) are 2 and 3. Correspondingly, substituting back into (x + y = 5), we get the pairs ((2, 3)) and ((3, 2)).Verifying these solutions in the original equations, both pairs satisfy (x^3 + y^3 = 35) and (x + y = 5).Therefore, the solutions are boxed{(2, 3)} and boxed{(3, 2)}.,<think>Okay, so I have this system of equations to solve. Let me write them down again to make sure I have them right:First equation: x³ + y³ = 35Second equation: x + y = 5Hmm, alright. The problem is to find the values of x and y that satisfy both equations. Let me think about how to approach this. I remember that when dealing with systems of equations, substitution is often a good method. Since the second equation is simpler, maybe I can solve for one variable in terms of the other and substitute it into the first equation. Let me try that.From the second equation, x + y = 5. If I solve for y, I get y = 5 - x. Then I can substitute this expression for y into the first equation. That way, the first equation will only have x's, and I can solve for x.So substituting y = 5 - x into the first equation gives:x³ + (5 - x)³ = 35Okay, let me expand (5 - x)³. Let me recall the formula for expanding (a - b)³. It's a³ - 3a²b + 3ab² - b³. So applying that here:(5 - x)³ = 5³ - 3*5²*x + 3*5*x² - x³Calculating each term:5³ = 1253*5²*x = 3*25*x = 75x3*5*x² = 15x²So putting it all together:(5 - x)³ = 125 - 75x + 15x² - x³Therefore, substituting back into the first equation:x³ + [125 - 75x + 15x² - x³] = 35Let me simplify this. The x³ and -x³ terms will cancel each other out. So:x³ - x³ + 15x² - 75x + 125 = 35Simplifying:15x² - 75x + 125 = 35Subtract 35 from both sides to set the equation to zero:15x² - 75x + 125 - 35 = 0Calculating 125 - 35: 90So:15x² - 75x + 90 = 0Hmm, okay. Let me check my math to make sure I didn't make a mistake. Let's go through the steps again:Original substitution:x³ + (5 - x)³ = 35Expanding (5 - x)³ gives 125 - 75x + 15x² - x³Adding x³ to that: x³ + 125 - 75x + 15x² - x³ = 35. The x³ terms do indeed cancel. Then 15x² -75x +125 =35. Subtract 35: 15x² -75x +90=0. That seems right.Now, let's simplify this quadratic equation. All coefficients are divisible by 15, so let's divide through by 15:(15x²)/15 - (75x)/15 + 90/15 = 0Which simplifies to:x² - 5x + 6 = 0Okay, that's a simpler quadratic equation. Let me factor this. Looking for two numbers that multiply to 6 and add up to -5. Wait, the equation is x² -5x +6. The factors would be (x - 2)(x - 3) because 2*3=6 and 2+3=5, but since the middle term is -5x, it should be (x - 2)(x - 3) = x² -5x +6. Yes, that works.So, factoring:(x - 2)(x - 3) = 0Therefore, the solutions are x = 2 and x = 3.Now, since we have x + y = 5, if x = 2, then y = 5 - 2 = 3. If x = 3, then y = 5 - 3 = 2.So the solutions are (2, 3) and (3, 2). Let me check these in the original equation to make sure.First, (x, y) = (2, 3):Check x³ + y³: 2³ + 3³ = 8 + 27 = 35. That matches the first equation. And x + y = 2 + 3 = 5. That matches the second equation.Next, (x, y) = (3, 2):Check x³ + y³: 3³ + 2³ = 27 + 8 = 35. Also matches. And x + y = 3 + 2 = 5. Correct again.So both solutions satisfy the system of equations. Therefore, the solutions are (2, 3) and (3, 2).Wait, but let me make sure there aren't any other solutions. The quadratic equation gave us two real roots, x=2 and x=3. Since the original system is a combination of a cubic and a linear equation, but through substitution, we ended up with a quadratic. Normally, a cubic equation can have up to three real roots, but in this case, substitution reduced it to a quadratic, so maximum two solutions. Therefore, these are the only solutions.Alternatively, maybe there's a way to use the identity for sum of cubes. Let me recall that x³ + y³ can be factored as (x + y)(x² - xy + y²). Maybe that's another approach. Let me try that.Given x³ + y³ =35 and x + y =5.Using the identity:x³ + y³ = (x + y)(x² - xy + y²) = 35We know x + y =5, so substituting:5*(x² - xy + y²) =35Divide both sides by 5:x² - xy + y² =7Now, we also know that x + y =5. Let me see if I can express x² + y² in terms of (x + y)².Recall that (x + y)² = x² + 2xy + y². Therefore, x² + y² = (x + y)² - 2xy = 25 - 2xy.So substituting back into x² - xy + y²:x² + y² - xy = (25 - 2xy) - xy =25 -3xySo we have 25 -3xy =7Subtract 25 from both sides:-3xy =7 -25= -18Divide both sides by -3:xy = (-18)/(-3) =6So xy=6Now we have x + y =5 and xy=6. So we can set up the quadratic equation whose roots are x and y. If x and y are roots of t² - (x + y)t + xy=0, then:t² -5t +6=0Which factors as (t -2)(t -3)=0, so t=2 and t=3. Therefore, the solutions are x=2,y=3 or x=3,y=2. Same as before.So both methods lead to the same solutions. This gives me more confidence that (2,3) and (3,2) are indeed the correct solutions. Therefore, the system has two real solutions.Just to make sure there isn't any other method that might give different solutions, let me think. For example, graphing. If I graph x + y =5, it's a straight line. The equation x³ + y³=35 is a cubic curve. The intersections should be at two points, which we found. Since both x and y are real numbers here, and the algebra shows two solutions, I don't think there are any other real solutions.Alternatively, maybe complex solutions, but the problem doesn't specify, so we can assume we're looking for real solutions.Therefore, the solutions are (2,3) and (3,2).Final AnswerThe solutions are boxed{(2, 3)} and boxed{(3, 2)}.</think>Given the system of equations:[left{begin{array}{l}x^{3} + y^{3} = 35 x + y = 5end{array}right.]First, solve the second equation for ( y ):[y = 5 - x]Substitute ( y = 5 - x ) into the first equation:[x^{3} + (5 - x)^{3} = 35]Expand ( (5 - x)^{3} ):[(5 - x)^{3} = 125 - 75x + 15x^{2} - x^{3}]Substitute back into the equation:[x^{3} + 125 - 75x + 15x^{2} - x^{3} = 35]Simplify:[15x^{2} - 75x + 125 = 35]Subtract 35 from both sides:[15x^{2} - 75x + 90 = 0]Divide by 15:[x^{2} - 5x + 6 = 0]Factor the quadratic equation:[(x - 2)(x - 3) = 0]Thus, the solutions for ( x ) are:[x = 2 quad text{or} quad x = 3]Using ( y = 5 - x ), we find the corresponding ( y ) values:- If ( x = 2 ), then ( y = 5 - 2 = 3 )- If ( x = 3 ), then ( y = 5 - 3 = 2 )Verify the solutions in both equations:- For ( (2, 3) ): - ( 2^{3} + 3^{3} = 8 + 27 = 35 ) - ( 2 + 3 = 5 )- For ( (3, 2) ): - ( 3^{3} + 2^{3} = 27 + 8 = 35 ) - ( 3 + 2 = 5 )Both solutions satisfy the system of equations. Therefore, the solutions are:[boxed{(2, 3)} quad text{and} quad boxed{(3, 2)}]]

🔑:## Solution.By the formula for the sum of cubes, we getbegin{aligned}& left{begin{array} { l } { ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 3 5 , } { x + y = 5 }end{array} Leftrightarrow left{begin{array}{l}(x+y) x+y=5end{array}left((x+y)^{2}-3 x yright)=35,right.right. & Leftrightarrowleft{begin{array} { l } { 5 ( 2 5 - 3 x y ) = 3 5 , } { x + y = 5 }end{array} Leftrightarrow left{begin{array} { l } { 2 5 - 3 x y = 7 , } { x + y = 5 }end{array} Leftrightarrow left{begin{array}{c}x y=6, x+y=5 .end{array}right.right.right.end{aligned}By Vieta's theorem, the possible solutions are; left{begin{array}{l}x_{1}=2, y_{1}=3end{array}right. or left{begin{array}{l}x_{2}=3, y_{2}=2 .end{array}right.Answer: (2 ; 3)(3 ; 2).

🧠:[<think>Okay, so I need to find the maximum and minimum values of the function y = sqrt(1 - x) + sqrt(x - 1/2). Then, once I have those, I need to compute a² + b², where a is the maximum and b is the minimum. Hmm, let's start by understanding the domain of this function because square roots have to be non-negative.First, let's look at the two square roots in the function. For sqrt(1 - x) to be real, the expression inside the square root must be greater than or equal to zero. So 1 - x ≥ 0 implies x ≤ 1. Similarly, for sqrt(x - 1/2), the inside expression must be ≥ 0, so x - 1/2 ≥ 0 implies x ≥ 1/2. Therefore, the domain of the function is all real numbers x such that 1/2 ≤ x ≤ 1. So x is between 0.5 and 1 inclusive.Now, the function is y = sqrt(1 - x) + sqrt(x - 1/2). To find the maximum and minimum values, maybe I can use calculus. That is, take the derivative, set it to zero, and find critical points. Alternatively, since it's a function of a single variable over a closed interval, the maximum and minimum must occur either at critical points or at the endpoints. So, maybe I can compute the derivative and see where it's zero or undefined, check those points, and compare with the endpoints.Alternatively, maybe there's a way to simplify this function or use some algebraic manipulation. Let me think. The function has two square roots. Let's denote:Let’s set t = x - 1/2. Then, since x is between 1/2 and 1, t will be between 0 and 1/2. Then, 1 - x = 1 - (t + 1/2) = 1/2 - t. So then, y = sqrt(1/2 - t) + sqrt(t). Hmm, that seems a bit more symmetric. So y = sqrt(1/2 - t) + sqrt(t), where t ∈ [0, 1/2]. Maybe this substitution helps.Alternatively, maybe we can square both sides to eliminate the square roots, but we have to be careful since squaring can introduce extraneous solutions. Let me try squaring y:y² = (sqrt(1 - x) + sqrt(x - 1/2))² = (1 - x) + (x - 1/2) + 2*sqrt{(1 - x)(x - 1/2)}. Let's compute this:1 - x + x - 1/2 = (1 - 1/2) + (-x + x) = 1/2. So y² = 1/2 + 2*sqrt{(1 - x)(x - 1/2)}. Therefore, y² = 1/2 + 2*sqrt{(1 - x)(x - 1/2)}.So, the square of y is equal to 1/2 plus twice the square root of (1 - x)(x - 1/2). Let me analyze the term inside the square root: (1 - x)(x - 1/2). Let's call this product Q = (1 - x)(x - 1/2). Let's see, when x is in [1/2, 1], then (1 - x) is in [0, 1/2] and (x - 1/2) is in [0, 1/2]. So both terms are non-negative, so Q is non-negative.Let me expand Q: (1 - x)(x - 1/2) = (1)(x) - (1)(1/2) - x(x) + x(1/2) = x - 1/2 - x² + (1/2)x = (x + 0.5x) - 1/2 - x² = 1.5x - x² - 0.5. Alternatively, perhaps it's better to write as quadratic in x:Q = (1 - x)(x - 1/2) = -x² + (3/2)x - 1/2. Let's verify:(1 - x)(x - 1/2) = 1*(x) - 1*(1/2) - x*(x) + x*(1/2) = x - 1/2 - x² + (1/2)x = (-x²) + (x + 0.5x) - 0.5 = -x² + 1.5x - 0.5. Yes, that's correct. So Q = -x² + (3/2)x - 1/2.Alternatively, we can write this quadratic in standard form. The quadratic is Q = -x² + (3/2)x - 1/2. Let's write it as Q = - [x² - (3/2)x + 1/2]. Completing the square inside the brackets:x² - (3/2)x + 1/2. The coefficient of x is -3/2, so half of that is -3/4, square is 9/16. So:x² - (3/2)x + 9/16 - 9/16 + 1/2 = (x - 3/4)^2 - 9/16 + 8/16 = (x - 3/4)^2 - 1/16.Therefore, Q = - [ (x - 3/4)^2 - 1/16 ] = - (x - 3/4)^2 + 1/16.So Q = - (x - 3/4)^2 + 1/16. Since (x - 3/4)^2 is always non-negative, the maximum value of Q is 1/16 when x = 3/4, and it decreases as x moves away from 3/4. The minimum value of Q occurs at the endpoints of the interval [1/2, 1]. Let's check:At x = 1/2: Q = (1 - 1/2)(1/2 - 1/2) = (1/2)(0) = 0.At x = 1: Q = (1 - 1)(1 - 1/2) = 0*(1/2) = 0.So Q reaches its maximum at x = 3/4, which is 1/16, and minimum 0 at the endpoints.Therefore, going back to y² = 1/2 + 2*sqrt(Q). Since Q ranges from 0 to 1/16, sqrt(Q) ranges from 0 to 1/4. Therefore, 2*sqrt(Q) ranges from 0 to 1/2. Hence, y² ranges from 1/2 + 0 = 1/2 to 1/2 + 1/2 = 1. Therefore, the square of y is between 1/2 and 1. Therefore, y ranges from sqrt(1/2) to 1. Hence, maximum y is 1, minimum y is sqrt(1/2).Wait, but let me check this again. If Q is between 0 and 1/16, then sqrt(Q) is between 0 and 1/4, so 2*sqrt(Q) is between 0 and 1/2. Therefore, adding to 1/2, y² is between 1/2 and 1. Therefore, y is between sqrt(1/2) and 1. So that would mean maximum value a = 1 and minimum value b = sqrt(1/2). Therefore, a² + b² = 1² + (sqrt(1/2))² = 1 + 1/2 = 3/2. So the answer would be 3/2, which is 1.5.But wait, let's verify this with actual values. Let's test x = 3/4. Then y = sqrt(1 - 3/4) + sqrt(3/4 - 1/2) = sqrt(1/4) + sqrt(1/4) = 1/2 + 1/2 = 1. So that's the maximum. Then, at x = 1/2, y = sqrt(1 - 1/2) + sqrt(1/2 - 1/2) = sqrt(1/2) + 0 = sqrt(1/2). Similarly, at x =1, y = sqrt(1 -1) + sqrt(1 -1/2) = 0 + sqrt(1/2) = sqrt(1/2). Wait, so the function reaches the maximum at x = 3/4 and the same minimum at both endpoints. Therefore, the maximum is 1 and the minimum is sqrt(1/2). Therefore, a =1, b= sqrt(1/2). Therefore, a² + b² =1 + 0.5=1.5, which is 3/2. So the answer should be 3/2. But let me check once more.Alternatively, maybe there's a mistake here. Let's see. The original substitution: when I set t =x -1/2, then x = t +1/2, so 1 -x =1 - t -1/2=1/2 -t. So y= sqrt(1/2 - t) + sqrt(t). Then, t is between 0 and 1/2. Let me write this as y(t)=sqrt(1/2 - t) + sqrt(t). Let's consider t ∈ [0,1/2]. Let's let t = (1/2) sin²θ or something? Maybe trigonometric substitution. Alternatively, use calculus here. Let's take the derivative of y with respect to t.dy/dt = (1/(2*sqrt(1/2 - t)))*(-1) + (1/(2*sqrt(t))).Set derivative to zero:-1/(2*sqrt(1/2 - t)) + 1/(2*sqrt(t)) =0Multiply both sides by 2:-1/sqrt(1/2 - t) + 1/sqrt(t) =0So, 1/sqrt(t) = 1/sqrt(1/2 - t)Therefore, sqrt(1/2 - t) = sqrt(t)Square both sides:1/2 - t = tSo, 1/2 = 2t => t=1/4So critical point at t=1/4, which is x=1/2 +1/4=3/4, which matches earlier result.Compute y at t=1/4: sqrt(1/2 -1/4) + sqrt(1/4)=sqrt(1/4)+sqrt(1/4)=1/2 +1/2=1. So maximum at 1, minimum at t=0 and t=1/2, which gives sqrt(1/2) +0 and 0 + sqrt(1/2). So yes, the minimum is sqrt(1/2). Therefore, a=1, b=sqrt(1/2). Then a² +b²=1 + 0.5=1.5=3/2. So the answer is 3/2.Wait, but let me check. If the function has maximum at 1 and minimum at sqrt(1/2), then the value a² +b² would be 1 + (sqrt(2)/2)^2=1 + (2/4)=1 +1/2=3/2. So yes, correct.But let me just confirm by another method. Let me use the Cauchy-Schwarz inequality or something else. Since the expression is y= sqrt(1 -x) + sqrt(x -1/2). Let me consider the two terms as vectors. Let me see, Cauchy-Schwarz says that (a + b)^2 ≤ (1 +1)(a² +b²). But maybe not helpful here.Alternatively, consider squaring the function. As done earlier, y²=1/2 + 2*sqrt{(1 -x)(x -1/2)}. Since the maximum of sqrt{(1 -x)(x -1/2)} is 1/4 (as found earlier), then maximum y²=1/2 + 2*(1/4)=1/2 +1/2=1. Similarly, minimum y² is 1/2 +0=1/2. Therefore, y ranges from sqrt(1/2) to 1. Therefore, the same conclusion. Therefore, a² +b²=1 +0.5=1.5. So the answer is 3/2. So 3/2 is 1.5, which is the value they are asking for.Therefore, the answer is 3/2, which in boxed form is boxed{dfrac{3}{2}}.Final Answerboxed{dfrac{3}{2}}</think>To find the maximum and minimum values of the function ( y = sqrt{1 - x} + sqrt{x - frac{1}{2}} ), we first determine the domain. The expressions inside the square roots must be non-negative, giving ( frac{1}{2} leq x leq 1 ).We then square the function to simplify the analysis:[y^2 = left( sqrt{1 - x} + sqrt{x - frac{1}{2}} right)^2 = (1 - x) + (x - frac{1}{2}) + 2 sqrt{(1 - x)(x - frac{1}{2})}]Simplifying the expression inside the square root:[(1 - x)(x - frac{1}{2}) = -x^2 + frac{3}{2}x - frac{1}{2}]This quadratic expression can be rewritten by completing the square:[-x^2 + frac{3}{2}x - frac{1}{2} = -left( x - frac{3}{4} right)^2 + frac{1}{16}]The maximum value of this quadratic expression is (frac{1}{16}) at ( x = frac{3}{4} ), and the minimum value is 0 at the endpoints ( x = frac{1}{2} ) and ( x = 1 ).Thus, the term ( 2 sqrt{(1 - x)(x - frac{1}{2})} ) ranges from 0 to ( frac{1}{2} ). Therefore, ( y^2 ) ranges from ( frac{1}{2} ) to ( 1 ), leading to ( y ) ranging from ( sqrt{frac{1}{2}} ) to ( 1 ).The maximum value ( a ) is ( 1 ) and the minimum value ( b ) is ( sqrt{frac{1}{2}} ). Calculating ( a^2 + b^2 ):[a^2 + b^2 = 1^2 + left( sqrt{frac{1}{2}} right)^2 = 1 + frac{1}{2} = frac{3}{2}]Thus, the value of ( a^2 + b^2 ) is [boxed{dfrac{3}{2}}],<think>Okay, so I need to find the maximum and minimum values of the function y = sqrt(1 - x) + sqrt(x - 1/2). Then, once I have those, called a and b respectively, I need to compute a squared plus b squared. Alright, let's start by understanding the domain of this function. Since there are square roots involved, the expressions inside them must be non-negative. First, sqrt(1 - x) requires that 1 - x ≥ 0, so x ≤ 1. Second, sqrt(x - 1/2) requires that x - 1/2 ≥ 0, so x ≥ 1/2. Therefore, the domain of the function is the interval [1/2, 1]. Got that. So x is between 0.5 and 1 inclusive. Now, the function is y = sqrt(1 - x) + sqrt(x - 1/2). To find its maximum and minimum values, I need to analyze how y behaves in this interval. Since it's a continuous function on a closed interval, by the Extreme Value Theorem, it should attain both a maximum and a minimum on this interval. Possible approaches could be taking the derivative and finding critical points, or maybe squaring the function to simplify it, but squaring might complicate things because of cross terms. Let's try taking the derivative first. Let me denote the function as y = (1 - x)^(1/2) + (x - 1/2)^(1/2). To find the derivative, I can apply the chain rule. The derivative dy/dx will be: - (1/2)(1 - x)^(-1/2) * (-1) + (1/2)(x - 1/2)^(-1/2) * (1)Simplifying that, the negatives and positives:(1/(2*sqrt(1 - x))) + (1/(2*sqrt(x - 1/2)))Wait, no. Wait, derivative of (1 - x)^(1/2) is (1/2)(1 - x)^(-1/2)*(-1) = -1/(2*sqrt(1 - x))And derivative of (x - 1/2)^(1/2) is (1/2)(x - 1/2)^(-1/2)*1 = 1/(2*sqrt(x - 1/2))Therefore, dy/dx = -1/(2*sqrt(1 - x)) + 1/(2*sqrt(x - 1/2))To find critical points, set dy/dx = 0:-1/(2*sqrt(1 - x)) + 1/(2*sqrt(x - 1/2)) = 0Multiply both sides by 2 to eliminate the denominators:-1/sqrt(1 - x) + 1/sqrt(x - 1/2) = 0Which gives 1/sqrt(x - 1/2) = 1/sqrt(1 - x)Take reciprocals on both sides:sqrt(x - 1/2) = sqrt(1 - x)Square both sides to eliminate the square roots:x - 1/2 = 1 - xBring variables to one side:x + x = 1 + 1/22x = 3/2x = 3/4So the critical point is at x = 3/4. Now we need to check if this is within the domain [1/2, 1]. 3/4 is 0.75, which is between 0.5 and 1, so yes. Therefore, the critical point is at x = 3/4. Now, to determine if this is a maximum or a minimum, we can check the second derivative or analyze the sign of the first derivative around x = 3/4. Alternatively, since we have a closed interval, we can just evaluate y at the critical point and at the endpoints, then compare. So let's compute y at x = 1/2, x = 3/4, and x = 1.First, at x = 1/2:y = sqrt(1 - 1/2) + sqrt(1/2 - 1/2) = sqrt(1/2) + sqrt(0) = sqrt(1/2) + 0 = 1/sqrt(2) ≈ 0.7071At x = 1:y = sqrt(1 - 1) + sqrt(1 - 1/2) = sqrt(0) + sqrt(1/2) = 0 + 1/sqrt(2) ≈ 0.7071At x = 3/4:Compute y:sqrt(1 - 3/4) + sqrt(3/4 - 1/2) = sqrt(1/4) + sqrt(1/4) = 1/2 + 1/2 = 1So, the function reaches 1 at x = 3/4 and 1/sqrt(2) ≈ 0.7071 at both endpoints. Therefore, the maximum value a is 1 and the minimum value b is 1/sqrt(2). But wait, the problem says "has a maximum value of a and a minimum value of b". Then compute a² + b². So a = 1, b = 1/sqrt(2). Therefore, a² + b² = 1 + (1/2) = 3/2. But wait, let me confirm if there are any other critical points. The derivative was calculated, and the only critical point was at x = 3/4, so the function is increasing from x = 1/2 to x = 3/4 and decreasing from x = 3/4 to x = 1. Therefore, the maximum is indeed at x = 3/4, and the minima are at the endpoints x = 1/2 and x = 1. But let me verify the behavior of the derivative. Let's pick a point between 1/2 and 3/4, say x = 0.6. Then compute dy/dx:-1/(2*sqrt(1 - 0.6)) + 1/(2*sqrt(0.6 - 0.5))Compute 1 - 0.6 = 0.4, sqrt(0.4) ≈ 0.6325, so first term: -1/(2*0.6325) ≈ -0.7906Second term: sqrt(0.6 - 0.5) = sqrt(0.1) ≈ 0.3162, so 1/(2*0.3162) ≈ 1.5811Therefore, dy/dx ≈ -0.7906 + 1.5811 ≈ 0.7905, which is positive. So the function is increasing from 0.5 to 0.75.Now, take a point between 0.75 and 1, say x = 0.8. Compute dy/dx:-1/(2*sqrt(1 - 0.8)) + 1/(2*sqrt(0.8 - 0.5))1 - 0.8 = 0.2, sqrt(0.2) ≈ 0.4472, so first term: -1/(2*0.4472) ≈ -1.118Second term: sqrt(0.3) ≈ 0.5477, so 1/(2*0.5477) ≈ 0.9129So dy/dx ≈ -1.118 + 0.9129 ≈ -0.205, which is negative. Therefore, the function is decreasing from 0.75 to 1.Thus, confirming that x = 3/4 is indeed the maximum. The endpoints at x = 1/2 and x = 1 give the minimum value.Therefore, the maximum value a is 1, and the minimum value b is 1/sqrt(2). Hence, a² + b² = 1² + (1/sqrt(2))² = 1 + 1/2 = 3/2. So the answer should be 3/2, which is 1.5, but in boxed form, as boxed{dfrac{3}{2}}.Wait, before I conclude, let me check if there's another way to approach this problem, maybe using Cauchy-Schwarz or some algebraic manipulation. Sometimes, these problems can be solved by squaring the function, but I need to check.Let me consider squaring y:y² = (sqrt(1 - x) + sqrt(x - 1/2))² = (1 - x) + (x - 1/2) + 2*sqrt{(1 - x)(x - 1/2)}.Simplify the terms:(1 - x) + (x - 1/2) = 1 - x + x - 1/2 = 1 - 1/2 = 1/2.Therefore, y² = 1/2 + 2*sqrt{(1 - x)(x - 1/2)}.So y² = 1/2 + 2*sqrt{(1 - x)(x - 1/2)}.So the maximum and minimum values of y correspond to the maximum and minimum of y², which in turn depends on the term sqrt{(1 - x)(x - 1/2)}. Let's denote t = (1 - x)(x - 1/2). Let's analyze this quadratic function.Let me expand t:(1 - x)(x - 1/2) = (1)(x) - (1)(1/2) - x(x) + x*(1/2) = x - 1/2 - x² + (x)/2 = -x² + (3x/2) - 1/2.But maybe it's easier to write as a quadratic function in x. Let me denote f(x) = (1 - x)(x - 1/2). Let's consider the interval x ∈ [1/2, 1].We can write f(x) = (1 - x)(x - 1/2). Let's find its maximum and minimum on [1/2, 1]. Since it's a quadratic function, which opens downward (as the coefficient of x² is -1), the maximum occurs at the vertex.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). Let's compute a and b from the expanded form:f(x) = -x² + (3/2)x - 1/2.So a = -1, b = 3/2.Vertex at x = - (3/2)/(2*(-1)) = (3/2)/2 = 3/4.So maximum value of f(x) is at x = 3/4:f(3/4) = (1 - 3/4)(3/4 - 1/2) = (1/4)(1/4) = 1/16.Therefore, the maximum value of t = (1 - x)(x - 1/2) is 1/16, occurring at x = 3/4, and the minimum value would be at the endpoints. Let's check:At x = 1/2: f(1/2) = (1 - 1/2)(1/2 - 1/2) = (1/2)(0) = 0.At x = 1: f(1) = (1 - 1)(1 - 1/2) = 0*(1/2) = 0.Therefore, t varies from 0 to 1/16. Therefore, sqrt(t) varies from 0 to 1/4.Hence, returning to y² = 1/2 + 2*sqrt(t). So when sqrt(t) is maximum (1/4), y² = 1/2 + 2*(1/4) = 1/2 + 1/2 = 1. Therefore, y = 1 (since y is non-negative). When sqrt(t) is minimum (0), y² = 1/2 + 0 = 1/2, so y = sqrt(1/2) = 1/√2. Therefore, this method also gives the maximum a = 1 and minimum b = 1/√2. Therefore, a² + b² = 1 + 1/2 = 3/2. So the answer is 3/2, which is consistent with the calculus approach.Therefore, regardless of the method, the answer is 3/2. Final Answerboxed{dfrac{3}{2}}</think>To find the maximum and minimum values of the function ( y = sqrt{1 - x} + sqrt{x - frac{1}{2}} ), we first determine the domain. The expressions inside the square roots must be non-negative, giving the interval ([ frac{1}{2}, 1 ]).Next, we find the critical points by taking the derivative of the function:[frac{dy}{dx} = -frac{1}{2sqrt{1 - x}} + frac{1}{2sqrt{x - frac{1}{2}}}]Setting the derivative to zero to find critical points:[-frac{1}{2sqrt{1 - x}} + frac{1}{2sqrt{x - frac{1}{2}}} = 0 implies sqrt{x - frac{1}{2}} = sqrt{1 - x}]Squaring both sides, we get:[x - frac{1}{2} = 1 - x implies 2x = frac{3}{2} implies x = frac{3}{4}]Evaluating the function at the critical point and endpoints:- At ( x = frac{1}{2} ): ( y = sqrt{frac{1}{2}} + 0 = frac{1}{sqrt{2}} )- At ( x = 1 ): ( y = 0 + sqrt{frac{1}{2}} = frac{1}{sqrt{2}} )- At ( x = frac{3}{4} ): ( y = sqrt{frac{1}{4}} + sqrt{frac{1}{4}} = frac{1}{2} + frac{1}{2} = 1 )Thus, the maximum value ( a ) is 1, and the minimum value ( b ) is ( frac{1}{sqrt{2}} ).To find ( a^2 + b^2 ):[a^2 + b^2 = 1^2 + left( frac{1}{sqrt{2}} right)^2 = 1 + frac{1}{2} = frac{3}{2}]The final answer is:[boxed{dfrac{3}{2}}]]

🔑:Solve: From 1-x geqslant 0, and x-frac{1}{2} geqslant 0, we getfrac{1}{2} leqslant x leqslant 1 text {. }Then y^{2}=frac{1}{2}+2 sqrt{-x^{2}+frac{3}{2} x-frac{1}{2}}=frac{1}{2}+2 sqrt{-left(x-frac{3}{4}right)^{2}+frac{1}{16}} text {. }Since frac{1}{2}<frac{3}{4}<1, therefore,when x=frac{3}{4}, y^{2} reaches its maximum value 1, so a=1;when x=frac{1}{2} or 1, y^{2} reaches its minimum value frac{1}{2}, so b=frac{sqrt{2}}{2}. Therefore, a^{2}+b^{2}=frac{3}{2}.

🧠:[<think>Alright, let's try to figure out this football tournament problem. So, we have n teams in a round-robin tournament. Each pair plays once. Wins give 2 points, losses 0, and draws give 1 point each. Three referees calculated the total points, getting three different totals: 3,086; 2,018; and 1,238. Then they averaged each pair of these numbers, and those averages were still different. Then they averaged all three initial numbers. So, there are seven numbers in total: the three original totals, the three pairwise averages, and the overall average. We know that one of these seven numbers is correct, and we need to find n, the number of teams.First, let me recall how total points work in a round-robin tournament. Since every match is between two teams, each match contributes either 2 points (if one team wins) or 2 points total (if it's a draw). Wait, hold on: if it's a win, 2 points to the winner; if it's a draw, 1 point each, so still 2 points total. Therefore, regardless of the number of draws or wins, the total points in the tournament should always be equal to the number of matches multiplied by 2.So, total points = 2 * number of matches. The number of matches in a round-robin tournament is C(n, 2) = n(n - 1)/2. Therefore, the correct total points should be 2 * [n(n - 1)/2] = n(n - 1). So, the correct total points must be n(n - 1).So, the key here is that among the seven numbers (three original totals, three pairwise averages, and the overall average), one of them must equal n(n - 1). So, our job is to check which of these seven numbers is equal to n(n - 1) for some integer n, and then verify the conditions given.First, let's list the seven numbers:Original totals: 3086, 2018, 1238.Pairwise averages:- (3086 + 2018)/2 = (5104)/2 = 2552- (3086 + 1238)/2 = (4324)/2 = 2162- (2018 + 1238)/2 = (3256)/2 = 1628Overall average: (3086 + 2018 + 1238)/3 = (6342)/3 = 2114So the seven numbers are: 3086, 2018, 1238, 2552, 2162, 1628, 2114. One of these is n(n - 1).So, we need to check which of these numbers can be written as n(n - 1) for some integer n. Let's check each:1. 3086: Let's solve n(n - 1) = 3086. Approximate n by sqrt(3086). sqrt(3086) is approx 55.55. So test n=56: 56*55=3080. Close, but 3080 vs 3086. Difference is 6. Not equal. n=55: 55*54=2970. Not close. So 3086 is not a possible total.2. 2018: Similarly, sqrt(2018) is approx 44.93. So test n=45: 45*44=1980. n=44: 44*43=1892. Neither is 2018. So not possible.3. 1238: sqrt(1238) ≈35.2. So n=35: 35*34=1190. n=36: 36*35=1260. 1260 vs 1238. Difference of 22. Not equal.4. 2552: sqrt(2552) ≈50.5. So n=51: 51*50=2550. Close to 2552, difference of 2. Not equal. n=50: 50*49=2450. No.5. 2162: sqrt(2162) ≈46.5. n=47: 47*46=2162. Exactly! So 47*46=2162. Let's check: 47*46 is indeed 2162. So this is possible. So n=47. Wait, 47 teams. Let me verify: 47*46=2162. Yes, 47*40=1880, 47*6=282; 1880+282=2162. Correct.6. 1628: sqrt(1628)≈40.35. Test n=40: 40*39=1560. n=41:41*40=1640. 1640 vs 1628. Difference of 12. Not equal.7. 2114: sqrt(2114)≈46. So n=46:46*45=2070. n=47:47*46=2162. Neither is 2114. So not possible.So, among all seven numbers, only 2162 can be expressed as n(n - 1), specifically n=47. Therefore, the correct total is 2162, which is the average of 3086 and 1238. Therefore, n=47.But wait, let me confirm if there could be any other numbers. Let me double-check each:3086: 56*55=3080, 57*56=3192. Not 3086.2018: 45*44=1980, 46*45=2070. Not 2018.1238: 35*34=1190, 36*35=1260. Not 1238.2552: 51*50=2550, 52*51=2652. Not 2552.2162: 47*46=2162. Correct.1628: 40*39=1560, 41*40=1640. Not 1628.2114: 46*45=2070, 47*46=2162. Not 2114.So only 2162 is a product of two consecutive integers. Therefore, n=47.But let's check if 2162 is indeed one of the seven numbers. Yes, it's the average of 3086 and 1238, which is (3086 + 1238)/2 = 2162. So the correct total is 2162, which is the average of two incorrect totals, but since one of the seven numbers is correct, and only 2162 fits n(n-1), then n=47.Hence, the answer is 47 teams.Final Answerboxed{47}</think>In a round-robin tournament with ( n ) teams, each pair of teams plays one match. The total points for all teams is always ( n(n-1) ) because each match contributes 2 points regardless of the result. The three referees reported totals of 3,086, 2,018, and 1,238. The pairwise averages and the overall average were calculated, resulting in seven numbers: 3,086; 2,018; 1,238; 2,552; 2,162; 1,628; and 2,114.We need to determine which of these numbers can be expressed as ( n(n-1) ). Checking each number:1. 3,086: Not a product of two consecutive integers.2. 2,018: Not a product of two consecutive integers.3. 1,238: Not a product of two consecutive integers.4. 2,552: Not a product of two consecutive integers.5. 2,162: ( 47 times 46 = 2,162 ), which fits the form ( n(n-1) ).6. 1,628: Not a product of two consecutive integers.7. 2,114: Not a product of two consecutive integers.The only number that fits is 2,162, which is the average of 3,086 and 1,238. Therefore, the correct total points is 2,162, leading to ( n = 47 ).[boxed{47}],<think>Okay, let me try to figure out this problem step by step. So, there's a round-robin football tournament with n teams. Each pair of teams plays one match. The scoring system is: 2 points for a win, 0 for a loss, and 1 point each for a draw. Three referees calculated the total points of all teams, and they got three different totals: 3,086; 2,018; and 1,238. Then they took the average of each pair of these numbers, and those averages were still different. After that, they averaged all three initial numbers. So in total, there are seven numbers: the original three totals, the three pairwise averages, and the overall average. Now, it's given that one of these seven numbers is correct. We need to find the number of teams n.First, let me recall that in a round-robin tournament, each pair of teams plays exactly once. So the total number of matches is C(n, 2) = n(n-1)/2. Now, for each match, depending on the result, the total points contributed to all teams is either 2 (if one team wins and the other loses) or 2 (if it's a draw, since each team gets 1 point). Wait, actually, in either case, the total points per match are 2. If it's a win/loss, 2 points go to the winner. If it's a draw, 1 point each, so also 2 points total. Therefore, regardless of the results, each match contributes exactly 2 points to the total.Therefore, the correct total points for all teams should be 2 * number of matches. Since number of matches is n(n-1)/2, the total points should be 2 * [n(n-1)/2] = n(n-1). So the correct total points is n(n-1). Now, the referees came up with three different totals: 3086, 2018, 1238. Then they averaged each pair, so the averages would be (3086+2018)/2, (3086+1238)/2, and (2018+1238)/2. Then they averaged all three, which would be (3086+2018+1238)/3. We know that one of these seven numbers (the three original totals, the three pairwise averages, or the overall average) is correct. So one of these numbers must equal n(n-1). Our job is to find which one of these numbers is equal to n(n-1), and thereby find n.So first, let's compute the pairwise averages and the overall average.First, let's list the three original totals: 3086, 2018, 1238.Pairwise averages:1. (3086 + 2018)/2 = (5104)/2 = 25522. (3086 + 1238)/2 = (4324)/2 = 21623. (2018 + 1238)/2 = (3256)/2 = 1628Then the overall average is (3086 + 2018 + 1238)/3 = (6342)/3 = 2114.So the seven numbers are:Original totals: 3086, 2018, 1238Pairwise averages: 2552, 2162, 1628Overall average: 2114Now, we need to check which of these numbers can be expressed as n(n-1) for some integer n. Let's check each one:First, check the original totals:3086: Let's see if n(n-1) = 3086. So solving n^2 - n - 3086 = 0. Using quadratic formula: n = [1 + sqrt(1 + 4*3086)]/2. Let's compute discriminant: 1 + 12344 = 12345. sqrt(12345) is approximately 111.11 (since 111^2=12321, 112^2=12544). So n is approx (1 + 111.11)/2 ≈ 56.05. Not an integer, so 3086 is not a possible total.2018: Similarly, n(n-1)=2018. Quadratic equation: n = [1 + sqrt(1 + 8072)]/2. sqrt(8073) is approx 89.85 (since 89^2=7921, 90^2=8100). So n ≈ (1 + 89.85)/2 ≈ 45.42. Not an integer.1238: n(n-1)=1238. Quadratic: n = [1 + sqrt(1 + 4952)]/2. sqrt(4953) ≈ 70.37 (since 70^2=4900, 71^2=5041). So n ≈ (1 + 70.37)/2 ≈ 35.68. Not an integer.So none of the original totals are integers when solving n(n-1). Therefore, the correct total must be among the pairwise averages or the overall average.Now check the pairwise averages:2552: n(n-1)=2552. Quadratic: n = [1 + sqrt(1 + 10208)]/2. sqrt(10209). Let's see: 100^2=10000, 101^2=10201, 102^2=10404. So sqrt(10209) is 101.0 (Wait, 101^2=10201, so 10201 vs 10209. Hmm, 101^2=10201, so sqrt(10209) is 101 + 8/(2*101) ≈ 101.04. So n ≈ (1 + 101.04)/2 ≈ 51.02. Not an integer.2162: n(n-1)=2162. Quadratic: n = [1 + sqrt(1 + 8648)]/2. sqrt(8649) = 93, because 93*93=8649. Therefore, discriminant is sqrt(1 + 8648)=sqrt(8649)=93. Therefore, n=(1 + 93)/2=94/2=47. So n=47. Let's check: 47*46=2162. Yes, 47*46=2162. So 2162 is a possible total. So n=47.Wait, but before jumping to conclusion, let's check other numbers.1628: n(n-1)=1628. Quadratic: n=[1 + sqrt(1 + 6512)]/2. sqrt(6513). Let's see: 80^2=6400, 81^2=6561. So sqrt(6513) is approx 80.7. So n≈ (1 + 80.7)/2≈40.85. Not integer.Overall average: 2114. Let's check n(n-1)=2114. Quadratic: n=[1 + sqrt(1 + 8456)]/2. sqrt(8457). 91^2=8281, 92^2=8464. So sqrt(8457)≈91.95. Therefore, n≈(1 +91.95)/2≈46.47. Not an integer.So among the seven numbers, only 2162 is equal to n(n-1) where n=47. Therefore, 2162 is the correct total, so n=47.But wait, let's make sure that 2162 is indeed one of the seven numbers. The pairwise averages were 2552, 2162, 1628. So yes, 2162 is there. So that's the correct total, hence n=47.Therefore, the answer should be 47.But let me just cross-verify once again. If n=47, total points would be 47*46=2162. So the total points should be 2162. Among the seven numbers, 2162 is present as one of the pairwise averages (average of 3086 and 1238). So since 2162 is correct, that's the number. Therefore, n=47.I think that's the solution.Final AnswerThe number of participating teams is boxed{47}.</think>In a round-robin tournament with ( n ) teams, each pair of teams plays exactly once, resulting in a total of ( binom{n}{2} = frac{n(n-1)}{2} ) matches. Each match contributes 2 points to the total score (regardless of a win or a draw), so the correct total points for all teams is ( n(n-1) ).Three referees calculated the total points as 3,086; 2,018; and 1,238. The pairwise averages of these totals are 2,552; 2,162; and 1,628. The overall average of the three totals is 2,114. Among these seven numbers (3,086; 2,018; 1,238; 2,552; 2,162; 1,628; 2,114), we need to find which one is equal to ( n(n-1) ).Checking each number:- The original totals (3,086; 2,018; 1,238) do not yield integer solutions for ( n ).- The pairwise averages (2,552; 2,162; 1,628) are checked: - For 2,162: Solving ( n(n-1) = 2162 ) gives ( n = frac{1 + sqrt{1 + 4 times 2162}}{2} = frac{1 + 93}{2} = 47 ), which is an integer.Thus, the correct total points is 2,162, corresponding to ( n = 47 ).[boxed{47}]]

🔑:(12) Sincebegin{array}{l}frac{3086+1238}{2}=2162, frac{3086+2018}{2}=2552, frac{1238+2018}{2}=1628, frac{3086+2018+1238}{3}=2114,end{array}n teams played a total of frac{n(n-1)}{2} matches, with each match accumulating 2 points, resulting in a total of n(n-1) points. Since the last digit of the product of two consecutive positive integers can only be 0, 2, 6, 2018, 1238, 1628, and 2114 are all excluded. From n(n-1)=2162, we get n=47; from n(n-1)=3086, since 56 times 55=3080, 57 times 56=3192, we get 56<n<57, which means there is no integer solution at this time; if n(n-1)=2552, since 50 times 51=2550, 51 times 52=2652, we have 51<n<52, which also has no integer solution. Therefore, only n=47.

🧠:[<think>Okay, let's see. I need to find the angle between the line of intersection of two planes and another plane that passes through three given points. Hmm. Alright, let me break this down step by step.First, the line of intersection of the two planes. The two planes are given by their equations: 2x - y - 3z + 5 = 0 and x + y - 2 = 0. To find the line of intersection, I remember that the direction vector of the line can be found by taking the cross product of the normal vectors of the two planes. That makes sense because the line is common to both planes, so its direction is perpendicular to both normals.So let me find the normal vectors first. For the first plane, 2x - y - 3z + 5 = 0, the coefficients of x, y, z are the normal components. So the normal vector n₁ is (2, -1, -3). For the second plane, x + y - 2 = 0, the normal vector n₂ is (1, 1, 0). Now, I need to compute the cross product of n₁ and n₂ to get the direction vector of the line. Let me recall the cross product formula: if n₁ = (a, b, c) and n₂ = (d, e, f), then n₁ × n₂ = (bf - ce, cd - af, ae - bd). Let me plug in the values here.n₁ is (2, -1, -3) and n₂ is (1, 1, 0). So the cross product components are:First component: (-1)(0) - (-3)(1) = 0 + 3 = 3Second component: (-3)(1) - (2)(0) = -3 - 0 = -3Third component: (2)(1) - (-1)(1) = 2 + 1 = 3Wait, is that right? Let me check again. The cross product formula is (n₁y*n₂z - n₁z*n₂y, n₁z*n₂x - n₁x*n₂z, n₁x*n₂y - n₁y*n₂x). So:First component: (-1)*0 - (-3)*1 = 0 + 3 = 3Second component: (-3)*1 - 2*0 = -3 - 0 = -3Third component: 2*1 - (-1)*1 = 2 + 1 = 3Yes, so the direction vector of the line of intersection is (3, -3, 3). Alternatively, we can write this as (1, -1, 1) by dividing each component by 3. That might make calculations easier later on. So direction vector v₁ = (1, -1, 1).Next, I need to find the equation of the plane passing through points M(-2, 0, 3), N(0, 2, 2), and K(3, -3, 1). To find the angle between the line and the plane, I think we need the angle between the line and the normal to the plane. Wait, no, the angle between a line and a plane is defined as the complement of the angle between the line and the normal to the plane. So if θ is the angle between the line and the normal, then the angle φ between the line and the plane is 90° - θ. So maybe I need to calculate θ first and then subtract it from 90°. But let's confirm that.Yes, I recall that the angle between a line and a plane is the angle between the line and its projection onto the plane, which is equal to 90° minus the angle between the line and the normal. So I need to find the angle between the direction vector of the line and the normal vector of the plane. Then take the sine of that angle as the angle between the line and the plane. Wait, maybe it's better to use the formula directly.The formula for the angle φ between a line with direction vector v and a plane with normal vector n is given by sin φ = |v · n| / (|v| |n|). Wait, no, that's the angle between the line and the normal. So φ = angle between line and plane, θ = angle between line and normal. Then sin φ = cos θ, because φ = 90° - θ. Therefore, sin φ = |v · n| / (|v| |n|). Wait, no, let me think again.If θ is the angle between the direction vector v and the normal vector n, then the angle between the line and the plane is 90° - θ. Therefore, sin(φ) = sin(90° - θ) = cos θ. But cos θ is |v · n| / (|v| |n|). Therefore, sin φ = |v · n| / (|v| |n|). So φ = arcsin(|v · n| / (|v| |n|)). So that's the formula we need.Alternatively, since φ = 90° - θ, and θ = angle between v and n, then φ = arcsin(|v · n| / (|v| |n|)). So maybe that's the way to go.But first, I need to find the normal vector of the plane passing through M, N, K. To find the normal vector of the plane, I can use the cross product of two vectors lying on the plane. Let's find two vectors on the plane. For example, vectors MN and MK.First, point M(-2, 0, 3), N(0, 2, 2), K(3, -3, 1).Vector MN is N - M = (0 - (-2), 2 - 0, 2 - 3) = (2, 2, -1)Vector MK is K - M = (3 - (-2), -3 - 0, 1 - 3) = (5, -3, -2)Then the normal vector n to the plane is MN × MK.Compute the cross product:MN = (2, 2, -1)MK = (5, -3, -2)Cross product components:i component: (2)(-2) - (-1)(-3) = -4 - 3 = -7j component: - [(2)(-2) - (-1)(5)] = - [ -4 + 5 ] = - [1] = -1k component: (2)(-3) - (2)(5) = -6 - 10 = -16Wait, wait. Let me recall the cross product formula. For vectors a = (a1, a2, a3) and b = (b1, b2, b3), the cross product is:(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)So let's compute MN × MK properly.MN = (2, 2, -1)MK = (5, -3, -2)So cross product is:i component: (2)(-2) - (-1)(-3) = (-4) - (3) = -7j component: (-1)(5) - (2)(-2) = (-5) - (-4) = -1Wait, wait, cross product formula is (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). So for j component, it's a3b1 - a1b3.Wait, let me check again. The cross product of (a1, a2, a3) × (b1, b2, b3) is:(i (a2b3 - a3b2), -j (a1b3 - a3b1), k (a1b2 - a2b1))So yes, the j component is negative of (a1b3 - a3b1)So for MN × MK:i component: a2b3 - a3b2 = (2)(-2) - (-1)(-3) = -4 - 3 = -7j component: - (a1b3 - a3b1) = - [ (2)(-2) - (-1)(5) ] = - [ -4 + 5 ] = - [1] = -1k component: a1b2 - a2b1 = (2)(-3) - (2)(5) = -6 -10 = -16Therefore, the cross product MN × MK is (-7, -1, -16). That is the normal vector of the plane. So n = (-7, -1, -16). Alternatively, we can write this as (7, 1, 16) since the direction is opposite, but since we need the normal vector for the angle calculation, the direction doesn't matter because we'll take the absolute value in the formula.So the normal vector of the plane is n = (-7, -1, -16). Let's keep it as is for now.Now, the direction vector of the line is v₁ = (1, -1, 1). As I found earlier. So now, we need to compute the angle between this direction vector and the normal vector of the plane. Then, take the arcsin of |v₁ · n| / (|v₁| |n|) to get the angle between the line and the plane.First, compute the dot product v₁ · n:v₁ · n = (1)(-7) + (-1)(-1) + (1)(-16) = -7 + 1 -16 = (-7 -16) +1 = -23 +1 = -22. So the absolute value is 22.Then, compute |v₁| = sqrt(1² + (-1)² +1²) = sqrt(1 +1 +1) = sqrt(3)Compute |n| = sqrt((-7)² + (-1)² + (-16)²) = sqrt(49 +1 +256) = sqrt(306). Hmm, 49 +1 is 50, 50 +256 is 306. So sqrt(306). Let me confirm that: 306 = 9*34, so sqrt(306) = 3*sqrt(34). Because 34 is 2*17, which is prime. So sqrt(306) = 3*sqrt(34). That might be useful later.So, sin φ = |v₁ · n| / (|v₁| |n|) = 22 / (sqrt(3)*sqrt(306)) = 22 / sqrt(3*306) = 22 / sqrt(918)Simplify sqrt(918). Let's see: 918 divided by 9 is 102. So sqrt(918) = 3*sqrt(102). Then 102 is 2*51 = 2*3*17. So sqrt(102) = sqrt(2*3*17). So sqrt(918) = 3*sqrt(102). Therefore, sin φ = 22 / (3*sqrt(102)).Wait, 3*102 is 306, so sqrt(306*3) = sqrt(918). Wait, but 3*306 is 918, so sqrt(918) is sqrt(3*306). Wait, but 3*306 = 918, yes. Alternatively, 3*306 = 918. So sqrt(3*306) = sqrt(918). Therefore, 22 / sqrt(918) is the same as 22 / (3*sqrt(102)) because 918 = 9*102. Wait, 918 divided by 9 is 102, so sqrt(918) = sqrt(9*102) = 3*sqrt(102). Therefore, sin φ = 22 / (3*sqrt(102)). So, sin φ = 22/(3*sqrt(102)). To rationalize the denominator, multiply numerator and denominator by sqrt(102):sin φ = (22 sqrt(102)) / (3*102) = (22 sqrt(102)) / 306. Simplify 22/306: divide numerator and denominator by 2: 11/153. So sin φ = (11 sqrt(102)) / 153. Hmm, but maybe it's better to leave it as 22/(3*sqrt(102)) or compute its decimal value.Alternatively, perhaps we can compute the angle directly using a calculator, but since the answer is probably expected to be in terms of inverse sine or in exact form, maybe we can just leave it as arcsin(22/(3*sqrt(102))) or simplify further.But let's check if there's a simplification. Let me compute 3*sqrt(102). sqrt(102) is approximately 10.0995, so 3*sqrt(102) ≈ 30.2985. Then 22 divided by that is approximately 0.726. So sin φ ≈ 0.726, which would mean φ ≈ arcsin(0.726) ≈ 46.6 degrees. But this is an approximate value. However, maybe the exact value is required. Alternatively, perhaps the answer can be presented in terms of an inverse trigonometric function. But let me check if the expression can be simplified.Alternatively, perhaps I made an error in calculation. Let me verify each step again.First, direction vector of the line of intersection: cross product of normals (2, -1, -3) and (1, 1, 0):Calculating cross product:|i j k||2 -1 -3||1 1 0|= i[(-1)(0) - (-3)(1)] - j[(2)(0) - (-3)(1)] + k[(2)(1) - (-1)(1)]= i[0 + 3] - j[0 + 3] + k[2 + 1]= 3i - 3j + 3k. So direction vector is (3, -3, 3), which simplifies to (1, -1, 1). Correct.Normal vector of the plane through M, N, K: cross product of MN and MK.MN is (2, 2, -1)MK is (5, -3, -2)Cross product MN × MK:|i j k||2 2 -1||5 -3 -2|= i[(2)(-2) - (-1)(-3)] - j[(2)(-2) - (-1)(5)] + k[(2)(-3) - (2)(5)]= i[-4 - 3] - j[-4 + 5] + k[-6 -10]= i(-7) - j(1) + k(-16)So (-7, -1, -16). Correct.Dot product of direction vector (1, -1, 1) and normal (-7, -1, -16):(1)(-7) + (-1)(-1) + (1)(-16) = -7 + 1 -16 = -22. Absolute value is 22. Correct.|v₁| = sqrt(1 +1 +1) = sqrt(3). Correct.|n| = sqrt(49 +1 +256) = sqrt(306). Correct.So sin φ = 22 / (sqrt(3)*sqrt(306)) = 22 / sqrt(918). Correct.But sqrt(918) can be written as sqrt(9*102) = 3*sqrt(102). So sin φ = 22/(3*sqrt(102)). Which simplifies to (22*sqrt(102))/(3*102) = (11*sqrt(102))/153. But perhaps it's better to leave it as is.Alternatively, maybe there's a mistake in the cross product for the normal vector. Let me check again.Wait, cross product MN × MK is:MN = (2, 2, -1)MK = (5, -3, -2)So cross product:i*(2*(-2) - (-1)*(-3)) - j*(2*(-2) - (-1)*5) + k*(2*(-3) - 2*5)= i*(-4 -3) - j*(-4 +5) + k*(-6 -10)= i*(-7) - j*(1) + k*(-16)So (-7, -1, -16). Correct. So that's the normal vector.Alternatively, maybe there's a different approach. For the plane passing through M, N, K, we could also find its equation and then find the normal vector from the equation.But since I already found the normal vector via cross product, which is correct, and the calculations seem right, then perhaps this is the right path.So the angle φ between the line and the plane is arcsin(22/(3*sqrt(102))). Alternatively, we can compute this angle in degrees.But let's check if this is the answer expected. Maybe we can rationalize or write it in terms of inverse sine. Alternatively, maybe there's a mistake in the direction vector or the normal vector.Wait, let me check the cross product again for MN and MK.MN is from M to N: (0 - (-2), 2 - 0, 2 - 3) = (2, 2, -1). Correct.MK is from M to K: (3 - (-2), -3 -0, 1 -3) = (5, -3, -2). Correct.So cross product components:i: 2*(-2) - (-1)*(-3) = -4 -3 = -7j: - [2*(-2) - (-1)*5 ] = - [ -4 +5 ] = -1k: 2*(-3) -2*5 = -6 -10 = -16Yes, so cross product is (-7, -1, -16). So normal vector is correct.Therefore, sin φ = 22/(3*sqrt(102)). Let me compute sqrt(102). 102 is 100 + 2, so sqrt(102) ≈ 10.0995. So 3*10.0995 ≈ 30.2985. Then 22 / 30.2985 ≈ 0.726. So arcsin(0.726) ≈ 46.6 degrees.But maybe the exact value is needed. Alternatively, maybe there's an error in the calculation of the angle.Wait, the angle between the line and the plane is φ = arcsin(|v · n| / (|v||n|)). So sin φ = |v · n| / (|v||n|). So yes, sin φ = 22 / (sqrt(3)*sqrt(306)).But sqrt(3)*sqrt(306) = sqrt(3*306) = sqrt(918). And 918 factors into 9*102, so sqrt(918)=3*sqrt(102). Therefore, sin φ = 22/(3*sqrt(102)) ≈ 22/(3*10.0995) ≈ 22/30.2985 ≈ 0.726. So arcsin(0.726) ≈ 46.6 degrees. So approximately 46.6 degrees. But I need to check if the problem expects an exact value or the answer in terms of inverse sine. Alternatively, maybe there's a miscalculation here.Alternatively, perhaps I can compute the angle using the direction vector and the normal vector. Wait, another way to calculate the angle between the line and the plane is to compute the angle between the line's direction vector and the plane's normal vector, then subtract from 90°. Let's try that.The angle θ between the direction vector v and the normal vector n is given by:cos θ = |v · n| / (|v||n|) = 22 / (sqrt(3)*sqrt(306)) ≈ 0.726. So θ ≈ arccos(0.726) ≈ 43.4 degrees. Therefore, the angle between the line and the plane is 90° - 43.4° ≈ 46.6°, which matches the previous result. So that's consistent. Therefore, the angle is approximately 46.6 degrees. But maybe the exact value is required, or perhaps expressed in terms of inverse trigonometric function.Alternatively, maybe there's a mistake in the direction vector of the line of intersection.Wait, direction vector is (1, -1, 1). Let me verify that again. The cross product of normals n₁ × n₂ = (2, -1, -3) × (1, 1, 0) = ( (-1)(0) - (-3)(1), (-3)(1) - (2)(0), (2)(1) - (-1)(1) ) = (0 +3, -3 -0, 2 +1) = (3, -3, 3). So direction vector is indeed (3, -3, 3) or simplified (1, -1, 1). Correct.Therefore, calculations seem correct. So the answer is arcsin(22/(3*sqrt(102))) or approximately 46.6 degrees.But since the problem says to put the final answer in boxed format, probably as an exact expression or a decimal. Maybe they want the exact value. Alternatively, let's check if there's a simplification.Wait, 22 and 3*sqrt(102) have no common factors. So the exact value is sin φ = 22/(3*sqrt(102)), so φ = arcsin(22/(3*sqrt(102))). Alternatively, rationalized, it is 22*sqrt(102)/(3*102) = 11*sqrt(102)/153, but that's probably not simpler.Alternatively, maybe there's a mistake in the cross product for the normal vector. Let me check once more.MN = (2, 2, -1)MK = (5, -3, -2)Cross product MN × MK:i component: (2)(-2) - (-1)(-3) = -4 -3 = -7j component: - [(2)(-2) - (-1)(5) ] = - [ -4 +5 ] = -1k component: (2)(-3) - (2)(5) = -6 -10 = -16Yes, all correct. So the normal vector is (-7, -1, -16). Correct.Alternatively, perhaps the problem expects the angle in radians? But likely in degrees. If we need to compute the exact value, it's not a standard angle, so perhaps we can leave it in terms of arcsin, but maybe the answer expects a decimal. However, since this is a math problem, sometimes exact forms are preferred.Wait, maybe I can rationalize the denominator in the sin φ expression.sin φ = 22/(3*sqrt(102)) = (22*sqrt(102))/(3*102) = (11*sqrt(102))/153. So this is equivalent.Alternatively, maybe the answer is expected to be expressed as the angle whose sine is 22/(3*sqrt(102)), so arcsin(22/(3√102)). But maybe the problem expects a simplified radical expression or a numerical answer. Since the problem doesn't specify, perhaps I should compute it numerically.Let me compute it step by step:First, compute sqrt(102) ≈ 10.0995Then 3*sqrt(102) ≈ 3*10.0995 ≈ 30.2985Then 22 / 30.2985 ≈ 0.726Then arcsin(0.726) ≈ 46.6 degrees. So approximately 46.6 degrees.Alternatively, using a calculator more precisely:Compute 22 divided by (3*sqrt(102)):sqrt(102) ≈ 10.099504943*10.09950494 ≈ 30.2985148222 / 30.29851482 ≈ 0.726109Then arcsin(0.726109) ≈ 46.66 degrees. So approximately 46.7 degrees.But perhaps the problem expects an exact answer, or maybe there is a miscalculation here. Let me check again.Wait, the problem says to find the angle between the line of intersection and the plane. So that angle is φ = arcsin(|v · n|/( |v||n| )).Wait, let me confirm the formula once more. Yes, angle between line and plane is defined as the angle between the line and its orthogonal projection onto the plane, which is equal to 90° minus the angle between the line and the normal. Therefore, sin φ = cos θ, where θ is the angle between the line and the normal. Alternatively, φ = 90° - θ. So using either formula is correct.But since I have v · n = -22, |v · n| = 22. Then |v| = sqrt(3), |n| = sqrt(306). So sin φ = 22/(sqrt(3)*sqrt(306)) = 22/sqrt(918). As 918 = 9*102, sqrt(918) = 3*sqrt(102). So sin φ = 22/(3*sqrt(102)) ≈ 22/(3*10.0995) ≈ 0.726. So φ ≈ arcsin(0.726) ≈ 46.7 degrees.Alternatively, maybe there is a mistake in the direction vector. Wait, if the direction vector was calculated incorrectly, that would affect the result. But I checked the cross product of the normals, which is correct.Alternatively, maybe I should present the answer as arcsin(22/(3*sqrt(102))) or in degrees. But since the problem doesn't specify, but in math problems like this, sometimes exact forms are expected, but I think in this case, since it's a numerical angle, it's better to approximate it. However, I need to check the instructions again. The problem says "put your final answer within boxed{}". It doesn't specify whether to leave it in terms of inverse sine or approximate. Given that, maybe it's better to leave it as an exact expression. Alternatively, perhaps there is a calculation error somewhere.Wait, let me compute the angle using another approach. Let's compute θ first, the angle between the direction vector and the normal, then subtract from 90° to get φ.cos θ = |v · n| / (|v| |n|) = 22 / (sqrt(3)*sqrt(306)) ≈ 0.726. So θ ≈ arccos(0.726) ≈ 43.4 degrees. Then φ = 90° - 43.4° ≈ 46.6°, which matches the previous result. So both methods give the same approximate angle.Alternatively, using more precise calculations:Compute 22 / (sqrt(3)*sqrt(306)) = 22 / sqrt(918) ≈ 22 / 30.2985 ≈ 0.726109Then θ = arccos(0.726109) ≈ 43.3312 degrees. Then φ = 90 - 43.3312 ≈ 46.6688 degrees. So approximately 46.67 degrees. So rounding to two decimal places, 46.67 degrees, or to one decimal, 46.7 degrees.Alternatively, maybe it's 47 degrees if rounded to the nearest degree. But the problem doesn't specify. But since in math problems like this, if an exact value isn't possible, an approximate decimal is acceptable, but maybe they want the exact expression.Alternatively, maybe there is a miscalculation here. Let me check if the plane equation is correctly found. If I can find the equation of the plane passing through M, N, K, then perhaps I can find the normal vector from there.Let me try to find the plane equation. Let's use the points M(-2, 0, 3), N(0, 2, 2), K(3, -3, 1).The general equation of a plane is ax + by + cz + d = 0. Substituting the three points into this equation:For M(-2, 0, 3): -2a + 0*b + 3c + d = 0 → -2a + 3c + d = 0For N(0, 2, 2): 0*a + 2b + 2c + d = 0 → 2b + 2c + d = 0For K(3, -3, 1): 3a -3b +1*c + d = 0 → 3a -3b + c + d = 0So we have three equations:1) -2a + 3c + d = 02) 2b + 2c + d = 03) 3a -3b + c + d = 0Let's solve this system.From equation 1: d = 2a - 3cSubstitute d into equation 2:2b + 2c + (2a -3c) = 0 → 2b + 2c + 2a -3c = 0 → 2a + 2b -c = 0 → equation 2': 2a + 2b - c = 0From equation 3: 3a -3b + c + (2a -3c) = 0 → 3a -3b + c + 2a -3c = 0 → 5a -3b -2c = 0 → equation 3': 5a -3b -2c = 0Now we have two equations:2a + 2b - c = 0 (2')5a -3b -2c = 0 (3')Let me solve these equations. Let's express c from equation 2':From equation 2': c = 2a + 2bSubstitute c = 2a + 2b into equation 3':5a -3b -2*(2a + 2b) = 0 → 5a -3b -4a -4b = 0 → (5a -4a) + (-3b -4b) = 0 → a -7b = 0 → a =7bSo a =7b. Then c =2a +2b =2*7b +2b=14b +2b=16bSo a=7b, c=16b. Then from equation 1: d=2a -3c =2*(7b) -3*(16b)=14b -48b= -34bThus, the plane equation is:ax + by + cz + d =0 →7b x + b y +16b z -34b =0We can factor out b (b ≠ 0):b(7x + y +16z -34)=0 →7x + y +16z -34=0So the equation of the plane is 7x + y +16z -34=0. Therefore, the normal vector is (7, 1, 16). Wait a minute! Earlier, when I calculated the cross product, I got (-7, -1, -16). But here, the normal vector is (7, 1, 16). That's just the negative of the previous one. Which makes sense, because cross product gives a normal vector, but its direction depends on the order of the vectors. So MN × MK gave (-7, -1, -16), which is equivalent to -(7, 1, 16). But since normal vectors can be in either direction, the plane equation here has normal vector (7, 1, 16). So, in the previous calculations, if I use (7, 1, 16) instead of (-7, -1, -16), the dot product would be:v₁ · n = (1, -1, 1) · (7, 1, 16) = 1*7 + (-1)*1 +1*16 = 7 -1 +16 = 22. Which is the same absolute value as before. Therefore, sin φ = 22/(sqrt(3)*sqrt(7² +1² +16²)) = 22/(sqrt(3)*sqrt(49 +1 +256)) = same as before, sqrt(306). So this confirms the previous result. So the normal vector is indeed (7, 1, 16), which is the same as the cross product result multiplied by -1. Therefore, the calculation is correct.Thus, the angle between the line of intersection and the plane is arcsin(22/(3*sqrt(102))) or approximately 46.7 degrees. Therefore, the answer is either expressed as an exact inverse sine or the approximate degree measure. Since the problem doesn't specify, but given the nature of the problem, I think the expected answer is the exact value in terms of inverse sine. However, in many geometry problems, angles are expected in degrees, and sometimes exact values are not possible, so an approximate answer may be acceptable.Alternatively, perhaps there's a mistake in the problem. For example, maybe the points are given with commas instead of semicolons, but in the problem statement, the points are given as M(-2 ; 0 ; 3), which is a bit unusual (semicolons instead of commas), but I took them as coordinates.Alternatively, maybe I can check using another method. For instance, find the parametric equations of the line of intersection and then compute the angle between that line and the plane. But that would lead to the same result.Alternatively, since the line's direction vector is (1, -1, 1) and the plane's normal vector is (7, 1, 16), the angle between them is θ with cosθ = |(1*7 + (-1)*1 +1*16)|/(sqrt(1+1+1)*sqrt(49+1+256)) = |7 -1 +16|/(sqrt(3)*sqrt(306)) = |22|/(sqrt(3)*sqrt(306)) = 22/(sqrt(3)*sqrt(306)), which is the same as before. Therefore, sinφ = 22/(sqrt(3)*sqrt(306)) as we found, leading to φ ≈ 46.7 degrees.Thus, I think the answer is approximately 46.7 degrees. So, I will box this as the final answer.Final Answerboxed{arcsinleft(dfrac{22}{3sqrt{102}}right)}Alternatively, if a numerical value is expected:boxed{46.7^circ}But since the problem might expect an exact answer, the first form is more appropriate.</think>To find the angle between the line of intersection of the planes (2x - y - 3z + 5 = 0) and (x + y - 2 = 0) and the plane passing through the points (M(-2, 0, 3)), (N(0, 2, 2)), and (K(3, -3, 1)), we follow these steps:1. Find the direction vector of the line of intersection: - The normal vectors of the planes are (mathbf{n_1} = (2, -1, -3)) and (mathbf{n_2} = (1, 1, 0)). - The direction vector of the line of intersection is the cross product of (mathbf{n_1}) and (mathbf{n_2}): [ mathbf{v_1} = mathbf{n_1} times mathbf{n_2} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 2 & -1 & -3 1 & 1 & 0 end{vmatrix} = (3, -3, 3) = (1, -1, 1) ]2. Find the normal vector of the plane passing through points (M), (N), and (K): - Vectors (overrightarrow{MN}) and (overrightarrow{MK}) are calculated as: [ overrightarrow{MN} = (2, 2, -1), quad overrightarrow{MK} = (5, -3, -2) ] - The normal vector (mathbf{n}) of the plane is the cross product of (overrightarrow{MN}) and (overrightarrow{MK}): [ mathbf{n} = overrightarrow{MN} times overrightarrow{MK} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 2 & 2 & -1 5 & -3 & -2 end{vmatrix} = (-7, -1, -16) ]3. Calculate the angle between the direction vector and the normal vector: - The dot product of (mathbf{v_1}) and (mathbf{n}): [ mathbf{v_1} cdot mathbf{n} = (1)(-7) + (-1)(-1) + (1)(-16) = -22 ] - Magnitudes of (mathbf{v_1}) and (mathbf{n}): [ |mathbf{v_1}| = sqrt{1^2 + (-1)^2 + 1^2} = sqrt{3}, quad |mathbf{n}| = sqrt{(-7)^2 + (-1)^2 + (-16)^2} = sqrt{306} ] - The sine of the angle (phi) between the line and the plane is: [ sin phi = frac{|mathbf{v_1} cdot mathbf{n}|}{|mathbf{v_1}| |mathbf{n}|} = frac{22}{sqrt{3} cdot sqrt{306}} = frac{22}{3sqrt{102}} ]Thus, the angle between the line of intersection and the plane is (boxed{arcsinleft(dfrac{22}{3sqrt{102}}right)}).,<think>Okay, so I need to find the angle between two things: first, the line of intersection of the planes 2x - y - 3z + 5 = 0 and x + y - 2 = 0, and second, the plane that passes through the points M(-2, 0, 3), N(0, 2, 2), and K(3, -3, 1). Hmm, angle between a line and a plane. I remember that the angle between a line and a plane is the complement of the angle between the line and the normal to the plane. So if θ is the angle between the line and the normal, then the angle we want is 90° - θ. Right?First, let me get the direction vector of the line of intersection of the two planes. For that, I can take the cross product of the normals of the two planes. The first plane has coefficients 2, -1, -3, so its normal vector is n1 = (2, -1, -3). The second plane is x + y - 2 = 0, so its normal vector is n2 = (1, 1, 0). The cross product of n1 and n2 will give the direction vector of the line of intersection.Calculating cross product:n1 × n2 = |i   j   k|         2   -1   -3         1    1   0= i*(-1*0 - (-3)*1) - j*(2*0 - (-3)*1) + k*(2*1 - (-1)*1)= i*(0 + 3) - j*(0 + 3) + k*(2 + 1)= 3i - 3j + 3kSo the direction vector of the line is (3, -3, 3). Maybe simplify it by dividing by 3? Then (1, -1, 1). That's easier.Now, the plane passing through points M, N, K. I need to find the equation of this plane first. To find the equation of a plane given three points, I can use the determinant method or find two vectors in the plane and take their cross product to get the normal vector.Let me find two vectors in the plane. Let's take vectors MN and MK.Point M is (-2, 0, 3), N is (0, 2, 2), K is (3, -3, 1).Vector MN = N - M = (0 - (-2), 2 - 0, 2 - 3) = (2, 2, -1)Vector MK = K - M = (3 - (-2), -3 - 0, 1 - 3) = (5, -3, -2)Now, the normal vector to the plane is the cross product of MN and MK.Calculating MN × MK:MN = (2, 2, -1)MK = (5, -3, -2)Cross product:i*(2*(-2) - (-1)*(-3)) - j*(2*(-2) - (-1)*5) + k*(2*(-3) - 2*5)Calculate each component:i: ( -4 - 3 ) = -7j: -( -4 - (-5) ) = - ( -4 + 5 ) = - (1) = -1k: ( -6 - 10 ) = -16Wait, let me check again:i component: MN_y * MK_z - MN_z * MK_y = 2*(-2) - (-1)*(-3) = -4 - 3 = -7j component: -(MN_x * MK_z - MN_z * MK_x) = -[2*(-2) - (-1)*5] = -[ -4 +5 ] = -[1] = -1k component: MN_x * MK_y - MN_y * MK_x = 2*(-3) - 2*5 = -6 -10 = -16So cross product vector is (-7, -1, -16). Let me write that as the normal vector n = (-7, -1, -16). Maybe we can multiply by -1 to make it positive, but direction doesn't matter for the normal.So the equation of the plane can be written as -7(x + 2) -1(y - 0) -16(z - 3) = 0? Wait, using point M(-2, 0, 3) to plug into the plane equation.Alternatively, general form: -7x - y -16z + D = 0. Substitute M(-2,0,3):-7*(-2) - 0 -16*3 + D = 0 → 14 -48 + D = 0 → -34 + D = 0 → D = 34So the plane equation is -7x - y -16z +34 =0. Maybe multiply both sides by -1 to make it nicer: 7x + y +16z -34 =0.Alright, so the normal vector of the second plane is (7, 1, 16). Wait, original cross product was (-7, -1, -16), so after multiplying by -1, it's (7,1,16). So the normal vector is n = (7, 1, 16).Now, to find the angle between the line (with direction vector v = (1, -1, 1)) and the plane (with normal vector n = (7,1,16)). As I thought earlier, the angle φ between the line and the plane is equal to 90° minus the angle θ between the line and the normal.First, find θ between v and n.The formula for the angle between two vectors is cosθ = (v ⋅ n)/(|v| |n|)Compute the dot product:v ⋅ n = 1*7 + (-1)*1 + 1*16 = 7 -1 +16 = 22Compute |v|: sqrt(1² + (-1)^2 +1²) = sqrt(1+1+1) = sqrt(3)Compute |n|: sqrt(7² +1² +16²) = sqrt(49 +1 +256) = sqrt(306). Let's see, 306 factors into 9*34, so sqrt(9*34)=3*sqrt(34). So |n|=3√34Thus, cosθ = 22 / (sqrt(3)*3*sqrt(34)) = 22 / (3*sqrt(102))Wait, sqrt(3)*sqrt(34) is sqrt(102), so yeah. So cosθ = 22/(3√102)But then θ = arccos(22/(3√102))But the angle we want is φ = 90° - θSince sinφ = cosθ (since sin(90° - θ) = cosθ), so sinφ = 22/(3√102)Therefore, φ = arcsin(22/(3√102))But maybe we can rationalize the denominator or simplify it.First, compute 3√102 ≈ 3*10.0995 ≈ 30.298522/30.2985 ≈ 0.726So sinφ ≈ 0.726, which would make φ ≈ arcsin(0.726) ≈ 46.6 degrees. But since the problem probably expects an exact answer, maybe in terms of inverse sine or cosine, but perhaps expressed in a simplified radical form?Wait, maybe I made a miscalculation in the normal vector. Let me double-check the cross product MN × MK.MN is (2,2,-1), MK is (5,-3,-2)Cross product:i: (2*(-2) - (-1)*(-3)) = (-4 -3) = -7j: -(2*(-2) - (-1)*5) = -(-4 +5) = -(1) = -1k: (2*(-3) -2*5) = (-6 -10) = -16So yes, cross product is (-7, -1, -16). So the normal vector is (-7, -1, -16) or (7,1,16). Either way, when taking the dot product, the direction doesn't matter for the angle. Let's confirm:If we use (7,1,16), then the dot product with v = (1,-1,1) is 7*1 +1*(-1) +16*1 =7 -1 +16=22. Same as before.So that's correct.Therefore, sinφ = 22/(3√102). To rationalize:Multiply numerator and denominator by √102:sinφ = (22√102)/(3*102) = (22√102)/306 = (11√102)/153 ≈ 0.726 as before.But maybe we can reduce 11 and 153. 153 ÷ 11 = 13.9, not integer. So probably simplest form is (11√102)/153.But the problem says to find the angle. Depending on the instructions, sometimes they want the answer in degrees, but since it's a math problem, they might expect the answer in terms of inverse sine or cosine. Alternatively, they might accept the exact value or a decimal. Wait, the original problem didn't specify, but given the variables, maybe they want an exact expression. But maybe there's a better way to compute.Alternatively, perhaps I made a mistake in computing the cross product. Let me check once again.MN vector: from M(-2,0,3) to N(0,2,2): (0 - (-2), 2 -0, 2 -3) = (2,2,-1). Correct.MK vector: from M(-2,0,3) to K(3,-3,1): (3 - (-2), -3 -0,1 -3) = (5,-3,-2). Correct.Cross product MN × MK:i component: (2*(-2) - (-1)*(-3)) = (-4 -3) = -7j component: -[(2*(-2) - (-1)*5)] = -[(-4 +5)] = -[1] = -1k component: (2*(-3) -2*5) = (-6 -10) = -16Yes, that's correct.So the normal vector is indeed (-7,-1,-16). So equation of the plane is -7x - y -16z + D =0. Plugging in M(-2,0,3):-7*(-2) -0 -16*3 +D = 14 -48 +D = -34 + D =0 → D=34. So plane equation: -7x - y -16z +34 =0 or 7x + y +16z =34.Normal vector (7,1,16). So all good.So, sinφ = 22/(3√102). Maybe simplifying √102. 102=2*51=2*3*17, so no square factors. So √102 is simplified. So sinφ = (22)/(3√102). To rationalize, multiply numerator and denominator by √102:sinφ = (22√102)/(3*102) = (11√102)/153 ≈ 0.726. So φ ≈ arcsin(0.726) ≈ 46.6 degrees.But perhaps we need to present the answer as arcsin(11√102/153) or in terms of arctangent? Wait, another way: The angle between the line and the plane can also be found by taking the angle between the line's direction vector and the plane's normal, then subtracting from 90. So, alternatively, if we have θ as the angle between the direction vector and the normal, then the angle between the line and the plane is 90 - θ, which is φ.Alternatively, since sinφ = cosθ. But maybe another formula.Wait, the formula for the angle between a line and a plane is given by φ = arcsin(|v ⋅ n| / (|v||n|)). Wait, no, that's the angle between the line and the normal. Wait, actually, the angle between the line and the plane is the complement of the angle between the line and the normal. So φ = 90° - θ where θ is the angle between the line and the normal. So sinφ = cosθ. But how does that relate to the formula?Wait, another approach: The angle between the line and the plane is the angle between the line and its projection onto the plane, which is equal to arcsin of the component of the direction vector perpendicular to the plane over the magnitude of the direction vector. Which would be arcsin(|v ⋅ n| / (|v||n|))). Wait, no, that's similar. Wait, let's clarify.If θ is the angle between v and n, then sinφ = |v ⋅ n| / (|v||n|). Wait, no, actually, since φ = 90° - θ, then sinφ = sin(90° - θ) = cosθ. And cosθ = |v ⋅ n|/(|v||n|). Therefore, sinφ = |v ⋅ n|/(|v||n|). Wait, but φ is the angle between the line and the plane, so it's between 0 and 90 degrees, so we take the absolute value.But in our case, the direction vector v and normal vector n. Since angle between line and plane is computed as arcsin of |v ⋅ n| / (|v||n| )? Wait, no:Wait, let's recall. The angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane, which is equal to 90° minus the angle between the line and the normal. So φ = 90° - θ, where θ is the angle between v and n.Therefore, sinφ = cosθ. Since cosθ = |v ⋅ n|/(|v||n|), so sinφ = |v ⋅ n|/(|v||n|). But since the angle between a line and a plane is always between 0° and 90°, we take the absolute value.Wait, but in our calculation earlier, we found sinφ = 22/(3√102) ≈0.726, which is positive, so φ≈46.6°. So the answer is arcsin(22/(3√102)) or arcsin(11√102/153). Hmm, but maybe simplifying 22/(3√102) as 11/(3√(102/4))? Wait, not sure. Alternatively, 22/3√102 can be written as 11/ ( (3/2)√102 ), but I don't think that's helpful.Alternatively, let's compute the numerical value.First, compute √102 ≈10.0995So 3√102 ≈3*10.0995≈30.298522/30.2985≈0.726Then, arcsin(0.726). Let me calculate that.Using calculator, sin(47°)= approx 0.731, sin(46°)=0.719. So 0.726 is between 46.6°, as I thought earlier. So approximately 46.6 degrees. But the problem probably expects an exact answer in terms of inverse sine, but I need to check if it's possible to express this angle more elegantly.Alternatively, maybe I made a mistake in calculation. Let me check again.The direction vector of the line: (1,-1,1). The normal vector of the plane: (7,1,16).Dot product: 1*7 + (-1)*1 +1*16 =7 -1 +16=22.|v|=sqrt(1+1+1)=sqrt(3)|n|=sqrt(49 +1 +256)=sqrt(306)=sqrt(9*34)=3*sqrt(34)Therefore, sinφ=22/(sqrt(3)*3*sqrt(34))=22/(3*sqrt(102))= same as before.So the answer is arcsin(22/(3√102)) or arcsin(11√102/153). Alternatively, rationalized.But maybe the problem wants the answer in degrees, but since it's a competition problem, perhaps they want an exact value. Alternatively, maybe there's a calculation mistake here.Wait, wait, angle between line and plane can be calculated as:φ = arcsin( |v ⋅ n| / (|v| |n|) )Wait, but wait, actually, the formula is:The sine of the angle between the line and the plane is equal to the absolute value of the dot product of the direction vector and the normal vector divided by the product of their magnitudes.But actually, no. Let me check.Wait, no. Wait, let me recall: The angle between a line and a plane is the angle between the line and its projection onto the plane, which is equal to φ = 90° - θ, where θ is the angle between the line and the normal. Therefore, sinφ = cosθ.But cosθ = |v ⋅ n| / (|v||n| )Therefore, sinφ = |v ⋅ n| / (|v||n| )Wait, so the formula is sinφ = |v ⋅ n| / (|v||n| )Therefore, in this case, sinφ = |22| / (sqrt(3)*sqrt(306)) )Wait, sqrt(3)*sqrt(306) = sqrt(3*306)=sqrt(918). Wait, but 306 is 9*34, so sqrt(306)=3*sqrt(34). Therefore, sqrt(3)*sqrt(306)=sqrt(3)*3*sqrt(34)=3*sqrt(102). So same as before. So sinφ = 22 / (3*sqrt(102)) ≈0.726. So yes, same result.Therefore, the angle is arcsin(22/(3√102)). Let me rationalize it as (22√102)/(3*102) = (11√102)/153. So arcsin(11√102/153). But this seems messy. Alternatively, maybe there is a miscalculation here.Alternatively, perhaps when calculating the cross product, the direction vector was miscalculated. Let me check cross product again.First plane: 2x - y -3z +5=0, normal vector (2,-1,-3)Second plane: x + y -2=0, normal vector (1,1,0)Cross product n1 × n2:|i   j   k|2   -1   -31   1   0= i*(-1*0 - (-3)*1) - j*(2*0 - (-3)*1) +k*(2*1 - (-1)*1)= i*(0 +3) - j*(0 +3) +k*(2 +1)=3i -3j +3k. So direction vector (3,-3,3) or simplified to (1,-1,1). Correct.So direction vector is (1,-1,1). Correct.Plane through M, N, K: cross product MN × MK was (-7, -1, -16). Normal vector (7,1,16). Correct.Dot product of direction vector and normal vector is 7*1 +1*(-1) +16*1=7 -1 +16=22. Correct.Magnitudes:|v|=sqrt(1 +1 +1)=sqrt(3)|n|=sqrt(49 +1 +256)=sqrt(306)=3*sqrt(34)Wait, 306=9*34. Therefore, sqrt(306)=3*sqrt(34). So |n|=3*sqrt(34)Therefore, sinφ=22/(sqrt(3)*3*sqrt(34))=22/(3*sqrt(102))But sqrt(102) cannot be simplified further. So yes, answer is arcsin(22/(3√102)).Alternatively, maybe convert into decimal degrees. Let's compute:First, sqrt(102)≈10.0995So 3*sqrt(102)≈30.298522 divided by 30.2985≈0.726Then, arcsin(0.726). Using calculator, arcsin(0.726)= approx 46.6 degrees.But perhaps the problem expects the answer in radians? Probably not, as it's more likely degrees unless specified.But since the problem is presented without specific units, maybe just present the exact value as arcsin(22/(3√102)) or simplified fraction with radicals. Alternatively, the answer might need to be written as arctan of something? Let me see.Alternatively, we can compute tanφ, since tanφ = sinφ / sqrt(1 - sin²φ )But that might complicate things. Alternatively, maybe there's a different approach.Wait, another way: The angle between a line and a plane can also be found using the formula:sinφ = |aA + bB + cC| / sqrt(a² + b² + c²) * sqrt(A² + B² + C²)Wait, no, that's similar to what we did. Alternatively, perhaps another approach.Alternatively, the angle between the line and the plane is equal to the angle between the line and its projection onto the plane. But I think the method we used is correct.So, unless there is an error in the calculation steps, the answer should be arcsin(22/(3√102)) which is approximately 46.6 degrees. But since mathematical problems often prefer exact forms, the answer would be expressed using the inverse sine function.Alternatively, maybe we can write the angle in terms of inverse tangent. Let's see.If sinφ = 22/(3√102), then tanφ = sinφ / sqrt(1 - sin²φ )Compute sinφ = 22/(3√102)So sin²φ = (484)/(9*102) = 484/918 = 242/459 ≈0.527sqrt(1 - sin²φ ) ≈sqrt(1 -0.527)=sqrt(0.473)≈0.688Thus, tanφ≈0.726 /0.688≈1.055Therefore, φ≈arctan(1.055)≈46.5 degrees. So same result. But this is not helpful for exact answer.Alternatively, maybe the problem expects the answer in degrees with decimal, but given that in olympiad problems, sometimes exact values are needed. However, since this involves square roots that don't simplify, perhaps leaving it in terms of arcsin is acceptable.Alternatively, maybe there's a miscalculation here, but I've checked each step twice. The direction vector is correct, normal vector is correct, dot product is correct, magnitudes are correct.Therefore, the angle is arcsin(22/(3√102)) which is approximately 46.6 degrees.Alternatively, is there a way to write this angle in exact terms? Let's compute 22/(3√102):Multiply numerator and denominator by sqrt(102):22*sqrt(102)/(3*102) = (11*sqrt(102))/153So sinφ = (11√102)/153, so φ = arcsin(11√102/153)Alternatively, rationalizing:But perhaps this is as simplified as it gets.Therefore, the answer is the angle whose sine is (11√102)/153, or approximately 46.6 degrees.Alternatively, the problem might expect the answer in degrees with exact expression, but since it's not a standard angle, perhaps just present the exact value in the box.Wait, the problem says "put your final answer within boxed{}". So maybe they expect the exact value expressed as an inverse sine, or the decimal approximation. But since the problem is about planes and lines, and given that the numbers don't lead to a nice angle, likely they expect the answer in terms of inverse trigonometric functions.But perhaps I can check if 22/(3√102) simplifies to something else.Wait, 22 and 3 and 102. 102=6*17. 22=2*11. So 22/(3√(6*17)) = (2*11)/(3√(6*17)) = not much simplification.Alternatively, if I write sqrt(102)=sqrt(2*3*17), but that's not helpful.Alternatively, perhaps the problem expects the answer as arccos(some expression), but since we have sinφ = 22/(3√102), then φ = arcsin(22/(3√102)). So that's the exact answer.Alternatively, in the Russian system, sometimes angles are presented in terms of fractions with π, but here it's impossible.Therefore, I think the answer is arcsin(22/(3√102)) or arcsin(11√102/153), but the problem might want the answer in degrees. But since no calculator is allowed, probably the exact answer is expected. Therefore, the angle is arcsin(22/(3√102)), which is the answer.Alternatively, if the answer needs to be written in a simplified form with a single square root, but I don't see a way.Alternatively, compute the angle in radians, but unlikely.Alternatively, maybe the answer is π/4 or something, but 22/(3√102)≈0.726, and arcsin(0.726)≈0.81 radians ≈46.6 degrees. So not a standard angle.Alternatively, maybe I made a mistake in the direction vector. Wait, the direction vector of the line of intersection is cross product of the two normals. Wait, the line of intersection lies on both planes, so its direction vector is perpendicular to both normals, hence cross product. So that part is correct.Alternatively, maybe there is an error in the plane equation. Let me check again.Points M(-2,0,3), N(0,2,2), K(3,-3,1).Vectors MN=(2,2,-1), MK=(5,-3,-2)Cross product MN × MK=(-7,-1,-16). So normal vector is (-7,-1,-16). So plane equation is -7x - y -16z + D=0. Plugging in M(-2,0,3):-7*(-2) -0 -16*3 + D=14 -48 +D= -34 + D=0 → D=34. So equation is -7x - y -16z +34=0 or 7x + y +16z=34. Correct.Normal vector is (7,1,16). Correct.Therefore, no mistake here.Therefore, I think the answer is as derived. So boxed answer would be boxed{arcsinleft( dfrac{22}{3sqrt{102}} right)}.Alternatively, if the problem expects the answer in degrees approximated to one decimal place, it would be approximately 46.6°, so boxed{46.6^circ}. But since mathematical problems often prefer exact forms, the arcsin expression is more likely. Alternatively, maybe the answer is expressed as the angle whose sine is 11√102 over 153, so boxed{arcsinleft( dfrac{11sqrt{102}}{153} right)}. Either form is acceptable, but likely the first one with 22/(3√102) is more straightforward.Final Answerboxed{arcsinleft( dfrac{22}{3sqrt{102}} right)}</think>To find the angle between the line of intersection of the planes (2x - y - 3z + 5 = 0) and (x + y - 2 = 0) and the plane passing through the points (M(-2, 0, 3)), (N(0, 2, 2)), and (K(3, -3, 1)), we follow these steps:1. Find the direction vector of the line of intersection of the two planes: - The normal vectors of the planes are (mathbf{n_1} = (2, -1, -3)) and (mathbf{n_2} = (1, 1, 0)). - The direction vector of the line of intersection is the cross product (mathbf{n_1} times mathbf{n_2}): [ mathbf{n_1} times mathbf{n_2} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 2 & -1 & -3 1 & 1 & 0 end{vmatrix} = 3mathbf{i} - 3mathbf{j} + 3mathbf{k} = (3, -3, 3) ] - Simplifying, the direction vector is (mathbf{v} = (1, -1, 1)).2. Find the equation of the plane passing through points (M), (N), and (K): - Vectors (overrightarrow{MN}) and (overrightarrow{MK}) are calculated as: [ overrightarrow{MN} = (2, 2, -1), quad overrightarrow{MK} = (5, -3, -2) ] - The normal vector to the plane is the cross product (overrightarrow{MN} times overrightarrow{MK}): [ overrightarrow{MN} times overrightarrow{MK} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 2 & 2 & -1 5 & -3 & -2 end{vmatrix} = -7mathbf{i} - mathbf{j} - 16mathbf{k} = (-7, -1, -16) ] - Using point (M(-2, 0, 3)), the plane equation is: [ -7(x + 2) - (y - 0) - 16(z - 3) = 0 implies 7x + y + 16z = 34 ] - The normal vector of this plane is (mathbf{n} = (7, 1, 16)).3. Calculate the angle between the line and the plane: - The angle (phi) between the line and the plane is (90^circ - theta), where (theta) is the angle between the direction vector (mathbf{v}) and the normal vector (mathbf{n}). - Using the dot product formula: [ costheta = frac{mathbf{v} cdot mathbf{n}}{|mathbf{v}| |mathbf{n}|} = frac{22}{sqrt{3} cdot 3sqrt{34}} = frac{22}{3sqrt{102}} ] - Therefore, (sinphi = costheta = frac{22}{3sqrt{102}}).The angle between the line of intersection and the plane is:[boxed{arcsinleft( dfrac{22}{3sqrt{102}} right)}]]

🔑:First, let's find the direction vector of the line of intersection of the given planes. For this, let x=t and solve the system of equations for y and z:left{begin{array}{l}y+3 z=2 t+5 y=-t+2end{array}right.We getx=t, y=2-t, z=1+tThus, the vector overrightarrow{boldsymbol{m}}=(1 ;-1 ; 1) is parallel to the line of intersection of the given planes. Next, we need to find a vector perpendicular to the plane passing through the points M(-2 ; 0 ; 3), N(0 ; 2 ; 2), and K(3 ;-3 ; 1). We will compute the coordinates of the vectors overrightarrow{M N} and overrightarrow{M K}:begin{aligned}& overrightarrow{M N}=(0-(-2) ; 2-0 ; 2-3)=(2 ; 2 ;-1) & overrightarrow{M K}=(3-(-2) ;-3-0 ; 1-3)=(5 ;-3 ;-2)end{aligned}Let vec{n}=(a ; b ; c) be a non-zero vector perpendicular to the desired plane. Then vec{n} cdot overrightarrow{M N}=0 and vec{n} cdot overrightarrow{M K}=0 orleft{begin{array}{l}2 a+2 b-c=0 5 a-3 b-2 c=0end{array}right.Multiply both sides of the first equation by -2 and add the result term by term to the second equation. We get the equation a-7 b=0. Let b=1. Then a=7, c=2 a+2 b=16. Let phi be the angle between the vectors vec{m}=(1 ;-1 ; 1) and vec{n}=(7 ; 1 ; 16). Thenbegin{aligned}& cos phi=frac{vec{m} cdot vec{m}}{|vec{m}| cdot|overrightarrow{mid vec{k}}|}=frac{1 cdot 7-1 cdot 1+1 cdot 16}{sqrt{1^{2}+1^{2}+1^{2} cdot sqrt{7^{2}+1^{2}+16^{2}}}}= & =frac{22}{sqrt{3 cdot 306}}=frac{22}{3 sqrt{102}}end{aligned}If alpha is the angle between the specified line and the plane M N K, thensin alpha=|cos phi|=frac{22}{3 sqrt{102}}## Answerarcsin frac{22}{3 sqrt{102}}