🧠:[<think>Okay, let's try to solve this problem. So we have a mapping f defined by f(x) = x - 2(x · a) a, where a is a constant vector. The condition is that f preserves the dot product, meaning f(x) · f(y) = x · y for any vectors x and y. We need to find which of the given options for the coordinates of a satisfy this condition.First, I need to recall what kind of transformation this mapping f represents. The formula x - 2(x · a) a looks familiar. If I remember correctly, this is the formula for a reflection over a line (in 2D) or a plane (in 3D) that is perpendicular to the vector a. So f is a reflection, right? And reflections are supposed to be linear transformations that preserve distances and angles, which would mean they preserve the dot product. But wait, maybe there's a catch here. The problem states that a is a constant vector, but does the reflection formula require a to be a unit vector?Let me check. The standard reflection formula across the hyperplane perpendicular to a vector n is given by:f(x) = x - 2 (x · n / ||n||²) nSo compared to our given formula, which is f(x) = x - 2(x · a) a, it seems like the standard formula would require n to be a unit vector, but in our case, the formula isn't divided by ||a||². Therefore, maybe the vector a in our problem needs to be a unit vector for f to be a reflection that preserves the dot product. Because if a is not a unit vector, then the scaling factor might mess up the preservation of the dot product.So, if that's the case, the condition we need is that ||a|| = 1. Then the formula simplifies to the standard reflection formula, which does preserve the dot product. Therefore, the coordinates of a must be a unit vector. So the answer is the option where the vector a has magnitude 1.Let me verify that. If a is a unit vector, then ||a||² = 1, so the reflection formula is indeed f(x) = x - 2(x · a) a, and since reflections preserve the dot product, the condition f(x) · f(y) = x · y holds. Therefore, the key is to check which of the given options have vectors a with magnitude 1.Let's compute the magnitude for each option:Option A: (√5/2, -1/2)Compute ||a||²: ( (√5/2)^2 + (-1/2)^2 ) = (5/4 + 1/4) = 6/4 = 3/2. So ||a|| = sqrt(3/2) ≈ 1.2247. Not 1.Option B: (√2/4, √2/4)||a||²: ( (√2/4)^2 + (√2/4)^2 ) = (2/16 + 2/16) = 4/16 = 1/4. So ||a|| = 1/2. Not 1.Option C: (3/4, 1/4)||a||²: (9/16 + 1/16) = 10/16 = 5/8. ||a|| ≈ 0.7906. Not 1.Option D: (-1/2, √3/2)||a||²: ( (-1/2)^2 + (√3/2)^2 ) = (1/4 + 3/4) = 1. So ||a|| = 1. That's a unit vector.Therefore, the answer should be D.But wait, let me think again. The problem states that f must satisfy f(x) · f(y) = x · y for any x, y. If a is not a unit vector, does this condition still hold?Suppose a is not a unit vector. Let's see. Let's compute f(x) · f(y):f(x) = x - 2(a · x)af(y) = y - 2(a · y)aThen f(x) · f(y) = [x - 2(a · x)a] · [y - 2(a · y)a]Expanding this:x · y - 2(a · x)(a · y) - 2(a · y)(x · a) + 4(a · x)(a · y)(a · a)Simplify:x · y - 4(a · x)(a · y) + 4(a · x)(a · y)(a · a)Factor out 4(a · x)(a · y):x · y + 4(a · x)(a · y)[ -1 + (a · a) ]So for f(x) · f(y) = x · y, the remaining terms must cancel out. That is:-4(a · x)(a · y) + 4(a · x)(a · y)(a · a) = 0Which simplifies to:4(a · x)(a · y)[ -1 + (a · a) ] = 0Since this must hold for any x and y, the coefficient must be zero. Therefore:-1 + (a · a) = 0 → a · a = 1 → ||a||² = 1Therefore, regardless of x and y, the equation holds if and only if ||a||² = 1, i.e., a is a unit vector.So my initial thought was correct. The vector a must be a unit vector. Therefore, among the options, only D has a vector with magnitude 1.So the answer is D.But just to be thorough, let me confirm each option's magnitude again:A: (sqrt(5)/2, -1/2)sqrt( (5/4) + (1/4) ) = sqrt(6/4) = sqrt(3/2) ≈ 1.2247. Not 1.B: (sqrt(2)/4, sqrt(2)/4)sqrt( (2/16) + (2/16) ) = sqrt(4/16) = sqrt(1/4) = 1/2. Not 1.C: (3/4, 1/4)sqrt(9/16 + 1/16) = sqrt(10/16) = sqrt(5/8) ≈ 0.7906. Not 1.D: (-1/2, sqrt(3)/2)sqrt(1/4 + 3/4) = sqrt(1) = 1. Yes, unit vector.Therefore, D is the correct answer.Final Answerboxed{D}</think>The mapping ( f ) is defined by ( f(x) = x - 2(x cdot a) a ), where ( a ) is a constant vector. We need to determine which coordinates of ( a ) satisfy the condition ( f(x) cdot f(y) = x cdot y ) for any ( x, y in A ).This mapping represents a reflection over a line perpendicular to ( a ). For ( f ) to preserve the dot product, ( a ) must be a unit vector. This is because the reflection formula requires the vector to be normalized (i.e., ( |a| = 1 )) to ensure the dot product is preserved.We check the magnitude of each given option for ( a ):- Option A: (left(frac{sqrt{5}}{2}, -frac{1}{2}right)) [ left|left(frac{sqrt{5}}{2}, -frac{1}{2}right)right|^2 = left(frac{sqrt{5}}{2}right)^2 + left(-frac{1}{2}right)^2 = frac{5}{4} + frac{1}{4} = frac{6}{4} = frac{3}{2} neq 1 ]- Option B: (left(frac{sqrt{2}}{4}, frac{sqrt{2}}{4}right)) [ left|left(frac{sqrt{2}}{4}, frac{sqrt{2}}{4}right)right|^2 = left(frac{sqrt{2}}{4}right)^2 + left(frac{sqrt{2}}{4}right)^2 = frac{2}{16} + frac{2}{16} = frac{4}{16} = frac{1}{4} neq 1 ]- Option C: (left(frac{3}{4}, frac{1}{4}right)) [ left|left(frac{3}{4}, frac{1}{4}right)right|^2 = left(frac{3}{4}right)^2 + left(frac{1}{4}right)^2 = frac{9}{16} + frac{1}{16} = frac{10}{16} = frac{5}{8} neq 1 ]- Option D: (left(-frac{1}{2}, frac{sqrt{3}}{2}right)) [ left|left(-frac{1}{2}, frac{sqrt{3}}{2}right)right|^2 = left(-frac{1}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2 = frac{1}{4} + frac{3}{4} = 1 ]Only Option D has a magnitude of 1, making it a unit vector. Therefore, the coordinates of ( a ) that satisfy the condition are given in Option D.[boxed{D}],<think>Alright, let's tackle this problem. So, we have a mapping f defined on planar vectors, right? The mapping is given by f(x) = x - 2(x · a)a, where a is a constant vector. The condition is that for any vectors x and y, the dot product of f(x) and f(y) must equal the dot product of x and y. The question is asking which of the given options for the coordinates of vector a satisfy this condition. The options are A, B, C, D. First, let me recall what this mapping f does. The expression x - 2(x · a)a looks familiar. I think this might be a reflection over a line in the plane. In linear algebra, the formula for reflecting a vector over a subspace involves subtracting twice the projection onto the subspace. If the subspace is a line (which, in the plane, it is), then the reflection formula is exactly x - 2(x · a)a, provided that a is a unit vector. Wait, is that right?Yes, the reflection of a vector x over a line can be written as x minus twice the projection of x onto the normal vector of the line. But wait, actually, if you're reflecting over a line, the formula is x - 2(x · a)a where a is a unit vector orthogonal to the line. Hmm, maybe I need to confirm that.Alternatively, if a is a unit vector, then the reflection across the line perpendicular to a would be x - 2(x · a)a. Wait, no. Let's think. If we have a line in the plane, and we want to reflect a vector over that line. The formula for the reflection can be written in terms of the projection onto the line. Alternatively, if the line is given by the direction of a unit vector u, then the reflection of x over the line is 2 Proj_u(x) - x. Alternatively, if we have a normal vector a to the line, then the reflection over the line (which is the same as reflecting through the hyperplane orthogonal to a) would be x - 2 (x · a / ||a||²) a. But in our case, the formula is x - 2(x · a) a. So, if a is a unit vector, then this would be x - 2(x · a)a, which is the reflection over the hyperplane orthogonal to a. But in the plane, reflecting over a hyperplane (a line) orthogonal to a would flip the component in the a direction. So yes, that makes sense. So, if a is a unit vector, then this mapping f is a reflection over the line orthogonal to a. And reflections are isometries, meaning they preserve distances and angles, so they should preserve the dot product. Therefore, if a is a unit vector, then f preserves the dot product. Conversely, if f preserves the dot product, then a must be a unit vector. Because if a is not a unit vector, then the scaling would affect the dot product. So, the key here is that a must be a unit vector. So, the coordinates of a must satisfy ||a|| = 1. Therefore, among the options, we need to check which of the vectors a have length 1.Let me check each option:Option A: (√5/2, -1/2)Compute the norm: ( (√5/2)^2 + (-1/2)^2 ) = (5/4 + 1/4) = 6/4 = 3/2. The square root of 3/2 is √(3/2), which is not 1. So, this vector has norm √(3/2) ≈ 1.2247, not 1. So, option A is not a unit vector.Option B: (√2/4, √2/4)Compute the norm: ( (√2/4)^2 + (√2/4)^2 ) = (2/16 + 2/16) = 4/16 = 1/4. The square root of 1/4 is 1/2. So, this vector has norm 1/2, which is not 1. So, option B is not a unit vector.Option C: (3/4, 1/4)Compute the norm: ( (3/4)^2 + (1/4)^2 ) = (9/16 + 1/16 ) = 10/16 = 5/8. The square root of 5/8 is √(5)/2√(2) ≈ 0.7906, which is not 1. So, option C is not a unit vector.Option D: (-1/2, √3/2)Compute the norm: ( (-1/2)^2 + (√3/2)^2 ) = (1/4 + 3/4 ) = 4/4 = 1. The square root of 1 is 1. So, this vector has norm 1. Therefore, option D is a unit vector.Therefore, the answer should be D.Wait, but let me verify again. The problem states that the mapping f satisfies f(x) · f(y) = x · y for any x, y. So, we need f to be an orthogonal transformation. As per the problem, f is given by f(x) = x - 2(x · a)a. So, this is a reflection if a is a unit vector. Reflections are orthogonal transformations, which preserve the dot product. Therefore, if a is a unit vector, then f is indeed an orthogonal transformation, and hence preserves the dot product. Conversely, if a is not a unit vector, then f would not preserve the dot product. Therefore, the necessary and sufficient condition is that a is a unit vector. Therefore, among the options, only D is a unit vector.Therefore, the correct answer is D.But just to make sure, let's test it with a specific example. Suppose we take a = (-1/2, √3/2). Since ||a|| = 1, as computed. Let's take an arbitrary vector x, say x = (1, 0). Then f(x) = x - 2(x · a) a. Compute x · a = (1)(-1/2) + (0)(√3/2) = -1/2. Therefore, f(x) = (1,0) - 2*(-1/2)*(-1/2, √3/2) = (1,0) - ( (-1/2)*(-1/2)*2, (-1/2)*(√3/2)*2 )Wait, let me compute step by step:x · a = -1/2.So, 2*(x · a)*a = 2*(-1/2)*(-1/2, √3/2) = (-1)*(-1/2, √3/2) = (1/2, -√3/2)Therefore, f(x) = x - 2*(x · a)*a = (1,0) - ( -2*(x · a)*a )? Wait, no. Wait, the formula is f(x) = x - 2*(x · a)*a. So, x · a is scalar, multiplied by 2, then multiplied by a. So, if x · a = -1/2, then 2*(x · a)*a = 2*(-1/2)*a = -1*a = -a. Therefore, f(x) = x - (-a) = x + a? Wait, that can't be. Wait, hold on.Wait, let's re-express:If x · a = -1/2, then 2*(x · a)*a = 2*(-1/2)*a = -a. Therefore, f(x) = x - (-a) = x + a.But that would mean f(x) = (1,0) + (-1/2, √3/2) = (1 - 1/2, 0 + √3/2) = (1/2, √3/2). Let's compute the norm of f(x). The original x is (1,0) with norm 1. f(x) is (1/2, √3/2). The norm is sqrt( (1/2)^2 + (√3/2)^2 ) = sqrt(1/4 + 3/4) = sqrt(1) = 1. So, the norm is preserved. Which is consistent with f being a reflection (which preserves norms). Similarly, take another vector y = (0,1). Compute f(y):y · a = (0)*(-1/2) + 1*(√3/2) = √3/2.Then 2*(y · a)*a = 2*(√3/2)*a = √3*a = √3*(-1/2, √3/2) = (-√3/2, (√3)^2 / 2 ) = (-√3/2, 3/2 )Therefore, f(y) = y - 2*(y · a)*a = (0,1) - (-√3/2, 3/2 ) = (0 + √3/2, 1 - 3/2 ) = (√3/2, -1/2 )Compute the norm of f(y): sqrt( (√3/2)^2 + (-1/2)^2 ) = sqrt( 3/4 + 1/4 ) = sqrt(1) = 1, which is the same as the norm of y. So, in both cases, the norm is preserved. Moreover, if we take the dot product of f(x) and f(y), which is (1/2, √3/2) · (√3/2, -1/2) = (1/2)(√3/2) + (√3/2)(-1/2) = (√3/4 - √3/4) = 0. The original x · y was (1,0) · (0,1) = 0, so that's preserved as well. So, at least in this case, the dot product is preserved.Therefore, since a is a unit vector, the mapping f is an isometry (reflection), which preserves the dot product. Hence, option D is correct. The other options are not unit vectors, so their corresponding mappings would not preserve the dot product.Therefore, the answer is D.Final Answerboxed{D}</think>The mapping ( f ) from the set ( A ) of planar vectors to ( A ) is defined by ( f(x) = x - 2(x cdot a)a ), where ( a ) is a constant vector. The condition given is that ( f(x) cdot f(y) = x cdot y ) for any ( x, y in A ). This implies that ( f ) must be an orthogonal transformation, specifically a reflection, which preserves the dot product. For ( f ) to be a reflection, ( a ) must be a unit vector.We need to check which of the given options for ( a ) have a norm of 1:- Option A: (left(frac{sqrt{5}}{2}, -frac{1}{2}right)) - Norm: (sqrt{left(frac{sqrt{5}}{2}right)^2 + left(-frac{1}{2}right)^2} = sqrt{frac{5}{4} + frac{1}{4}} = sqrt{frac{6}{4}} = sqrt{frac{3}{2}} neq 1)- Option B: (left(frac{sqrt{2}}{4}, frac{sqrt{2}}{4}right)) - Norm: (sqrt{left(frac{sqrt{2}}{4}right)^2 + left(frac{sqrt{2}}{4}right)^2} = sqrt{frac{2}{16} + frac{2}{16}} = sqrt{frac{4}{16}} = sqrt{frac{1}{4}} = frac{1}{2} neq 1)- Option C: (left(frac{3}{4}, frac{1}{4}right)) - Norm: (sqrt{left(frac{3}{4}right)^2 + left(frac{1}{4}right)^2} = sqrt{frac{9}{16} + frac{1}{16}} = sqrt{frac{10}{16}} = sqrt{frac{5}{8}} neq 1)- Option D: (left(-frac{1}{2}, frac{sqrt{3}}{2}right)) - Norm: (sqrt{left(-frac{1}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{3}{4}} = sqrt{1} = 1)Only option D has a norm of 1, making it a unit vector. Therefore, the coordinates of ( a ) must be (left(-frac{1}{2}, frac{sqrt{3}}{2}right)).(boxed{D})]

🔑:begin{array}{l}text {-,1.D. } f(x) cdot f(y)=[x-2(x cdot a) a] cdot[y-2(y cdot a) a] =x cdot y+4left(a^{2}-1right)(x cdot a)(y cdot a)end{array}Therefore, from the condition, we should get a^{2}-1=0, i.e., |a|=1.

🧠:[<think>Okay, so I need to find the length of side a in triangle ABC. Let me recall what information I have: angle A is 60 degrees, the side opposite to angle B is 2 (so b = 2), and the area of the triangle is 2√3. Hmm, how do I approach this?First, the area of a triangle can be calculated using the formula: (1/2)*ab*sin(C), where a and b are two sides and C is the included angle between them. But wait, in this case, angle A is 60 degrees, and sides a and b are opposite angles A and B. So if I consider sides b and c with angle A between them, maybe? Wait, maybe I need to clarify which sides are adjacent to angle A.In triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. So angle A is between sides b and c. So if I want to use the area formula with angle A, I need sides b and c. The formula would be Area = (1/2)*b*c*sin(A). But I know the area is 2√3, and angle A is 60 degrees, so sin(60°) is √3/2. Let me write that down:Area = (1/2)*b*c*sin(60°) = 2√3Plugging in the known values:(1/2)*2*c*(√3/2) = 2√3Simplify the left side:(1/2)*2*c*(√3/2) = (1)*c*(√3/2) = (c√3)/2Set that equal to 2√3:(c√3)/2 = 2√3Multiply both sides by 2:c√3 = 4√3Divide both sides by √3:c = 4Okay, so I found that side c is 4. Now, I need to find side a. Since I have sides b and c, and angle A between them, maybe I can use the Law of Cosines to find side a. The Law of Cosines states that a² = b² + c² - 2bc*cos(A). Let's apply that.Given that angle A is 60°, b = 2, c = 4:a² = 2² + 4² - 2*2*4*cos(60°)Calculate each term:2² = 44² = 162*2*4 = 16cos(60°) = 0.5, so 16*0.5 = 8So putting it all together:a² = 4 + 16 - 8 = 20 - 8 = 12Therefore, a = √12 = 2√3Wait, so the length of side a is 2√3. Let me verify if this makes sense.Alternatively, maybe I can use the Law of Sines to check. The Law of Sines states that a/sin(A) = b/sin(B) = c/sin(C). But I don't know angles B or C yet. Maybe I can find angle B or C first?Since I found sides b = 2, c = 4, and angle A = 60°, maybe I can use the Law of Sines to find another angle. Let's try to find angle B.Using Law of Sines: a/sin(A) = b/sin(B)We already found a = 2√3, so:2√3 / sin(60°) = 2 / sin(B)Calculate sin(60°) = √3/2, so:2√3 / (√3/2) = 2 / sin(B)Simplify left side:2√3 * (2/√3) = 4So 4 = 2 / sin(B)Thus, sin(B) = 2/4 = 1/2Therefore, angle B is arcsin(1/2), which is 30° or 150°. But in a triangle, the sum of angles is 180°, so if angle A is 60°, angle B can't be 150°, because 60 + 150 = 210 > 180. Therefore, angle B must be 30°.Then angle C would be 180° - 60° - 30° = 90°. Let me check if that makes sense. If angle C is 90°, then side c would be the hypotenuse. Using Pythagoras theorem, a² + b² = c²?But wait, in this case, if angle C is 90°, then sides a and b would be the legs, and c the hypotenuse. But we have a = 2√3, b = 2, and c = 4.Check if (2√3)² + 2² = 4²:(4*3) + 4 = 12 + 4 = 164² = 16. Yes, it works. So angle C is 90°, which aligns with our earlier conclusion. So that seems consistent.Therefore, using both the Law of Cosines and Law of Sines, and verifying with the Pythagorean theorem, the value of a is indeed 2√3. Let me recap the steps to make sure I didn't skip anything.1. Used the area formula with sides b and c and angle A to find c.2. Applied the Law of Cosines with sides b, c, and angle A to find a.3. Verified using Law of Sines and checked angles to ensure consistency.4. Confirmed with Pythagoras theorem since angle C turned out to be 90°, making the triangle a right-angled one.Everything checks out. So I think the answer is a = 2√3.Final AnswerThe value of ( a ) is boxed{2sqrt{3}}.</think>Given a triangle ( triangle ABC ) with ( A = 60^circ ), ( b = 2 ), and the area ( S_{triangle ABC} = 2sqrt{3} ), we need to find the value of ( a ).1. Calculate side ( c ) using the area formula: [ text{Area} = frac{1}{2} cdot b cdot c cdot sin(A) ] Substituting the given values: [ 2sqrt{3} = frac{1}{2} cdot 2 cdot c cdot sin(60^circ) ] Simplifying with ( sin(60^circ) = frac{sqrt{3}}{2} ): [ 2sqrt{3} = frac{1}{2} cdot 2 cdot c cdot frac{sqrt{3}}{2} ] [ 2sqrt{3} = frac{csqrt{3}}{2} ] Solving for ( c ): [ csqrt{3} = 4sqrt{3} ] [ c = 4 ]2. Apply the Law of Cosines to find ( a ): [ a^2 = b^2 + c^2 - 2bc cdot cos(A) ] Substituting ( b = 2 ), ( c = 4 ), and ( cos(60^circ) = 0.5 ): [ a^2 = 2^2 + 4^2 - 2 cdot 2 cdot 4 cdot 0.5 ] [ a^2 = 4 + 16 - 8 ] [ a^2 = 12 ] [ a = sqrt{12} = 2sqrt{3} ]3. Verification using the Law of Sines and Pythagorean theorem: - Using the Law of Sines, we found angle ( B = 30^circ ) and angle ( C = 90^circ ). - Verified with the Pythagorean theorem confirming ( a = 2sqrt{3} ), ( b = 2 ), and ( c = 4 ) form a right triangle.Thus, the value of ( a ) is boxed{2sqrt{3}}.,<think>Okay, so I need to find the length of side a in triangle ABC, where angle A is 60 degrees, side b is 2 units long, and the area of the triangle is 2√3. Let me recall the formulas related to triangles that might help here. First, I know that the area of a triangle can be calculated using the formula: [ text{Area} = frac{1}{2}ab sin C ]But in this case, angle A is given, and sides b and a are mentioned. Wait, actually, the formula should be using two sides and the included angle. Since angle A is 60 degrees, and the sides adjacent to angle A are b and c, right? Wait, no. Wait, angle A is opposite side a. So sides b and c are the other two sides. Hmm. Maybe I should draw a quick sketch to visualize.Let me label the triangle: angle A is 60 degrees, opposite side a. Sides b and c are adjacent to angle A? Wait, no. In a triangle, each angle is opposite a side. So angle A is opposite side a, angle B opposite side b, and angle C opposite side c. So sides b and c are adjacent to angle A. So if I want to use the area formula with angle A, I need sides b and c, because those are the sides adjacent to angle A. The formula would be:[ text{Area} = frac{1}{2} times b times c times sin A ]Yes, that makes sense. Because angle A is between sides b and c. So plugging in the values we have:Area = 2√3 = (1/2)*b*c*sin(60°)We know b = 2, so we can solve for c.Let me write that equation out:[ 2sqrt{3} = frac{1}{2} times 2 times c times sin 60° ]Simplify the right side:First, 1/2 * 2 = 1, so we get:[ 2sqrt{3} = c times sin 60° ]Sin 60° is √3/2, so:[ 2sqrt{3} = c times (sqrt{3}/2) ]To solve for c, multiply both sides by 2/√3:[ c = 2sqrt{3} times (2/sqrt{3}) ]Simplify:The √3 cancels out, and 2*2 =4, so c =4.Okay, so now I know that side c is 4. So now the triangle has sides a (opposite 60°), b=2, and c=4. Now I need to find a. Maybe I can use the Law of Cosines here? Since I know angle A and the two sides adjacent to it (b and c). Wait, Law of Cosines states that:[ a^2 = b^2 + c^2 - 2bc cos A ]Yes, perfect. Let's plug in the known values.Given angle A is 60°, so cos 60° is 0.5. So:[ a^2 = 2^2 + 4^2 - 2 times 2 times 4 times 0.5 ]Compute each term:2^2 =44^2=162*2*4=16, multiplied by 0.5 is 8.So:[ a^2 = 4 + 16 - 8 ]Which is:[ a^2 = 20 - 8 = 12 ]So a = √12 = 2√3.Wait, let me check if that makes sense. Alternatively, maybe I could use the Law of Sines. Let me verify with that.Law of Sines states that:[ frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ]We know a, angle A, b, and c. But since we found a, maybe we can check if the ratios hold. Let's see:We have a = 2√3, angle A = 60°, so sin A = √3/2. So:[ frac{2√3}{√3/2} = frac{2√3 * 2}{√3} = frac{4√3}{√3} = 4 ]So the ratio for a/sin A is 4. Then, check for b/sin B. We know b=2, so:2/sin B = 4 => sin B = 2/4 = 1/2. Therefore, angle B = 30° or 150°. But since the sum of angles in a triangle is 180°, and angle A is 60°, if angle B were 150°, then angle C would be negative, which is impossible. So angle B must be 30°, making angle C = 90°.Wait, angle C would be 180 - 60 - 30 = 90°. So angle C is 90°, so triangle is right-angled at C. Then, sides a, b, c would form a right triangle with legs b=2, c=4, and hypotenuse a. Wait, but in a right-angled triangle, the hypotenuse should be the longest side. Here, c=4 is longer than a=2√3 (~3.464), so that's correct. So hypotenuse is c=4, but wait, no. Wait, if angle C is 90°, then the hypotenuse is opposite angle C, which is side c. Wait, but side c is 4. So hypotenuse is 4, and the legs are a and b. Wait, but if angle C is 90°, then sides a and b are the legs. But according to our previous calculation, side a is 2√3, side b is 2, so the legs are 2 and 2√3, hypotenuse 4. Let's check with Pythagoras:[ (2)^2 + (2√3)^2 = 4 + 12 = 16 = 4^2 ]Yes, that works. So the triangle is indeed right-angled at C. So everything checks out.Alternatively, maybe I could have approached the problem using the area formula with sides and angles. But in any case, the answer seems to be 2√3.Wait, let me recap to make sure I didn't skip any steps. Given angle A = 60°, sides b = 2, area = 2√3. Calculated side c using area formula, got c=4. Then used Law of Cosines to find a, got 2√3. Checked with Law of Sines and Pythagorean theorem, everything consistent. So I think that's correct.Another way to cross-validate: If the triangle is right-angled at C, then area should also be (1/2)*a*b. Since legs are a and b. So:Area = (1/2)*a*b = (1/2)*(2√3)*2 = (1/2)*4√3 = 2√3. Which matches the given area. So that's correct.Therefore, the value of a is 2√3.Final AnswerThe value of ( a ) is boxed{2sqrt{3}}.</think>Given (triangle ABC) with (A = 60^circ), (b = 2), and the area (S_{triangle ABC} = 2sqrt{3}), we need to find the value of (a).First, we use the area formula for a triangle involving two sides and the included angle:[text{Area} = frac{1}{2} times b times c times sin A]Substituting the given values:[2sqrt{3} = frac{1}{2} times 2 times c times sin 60^circ]Simplifying, we know (sin 60^circ = frac{sqrt{3}}{2}):[2sqrt{3} = 1 times c times frac{sqrt{3}}{2}]Solving for (c):[2sqrt{3} = c times frac{sqrt{3}}{2} implies c = frac{2sqrt{3} times 2}{sqrt{3}} = 4]Next, we use the Law of Cosines to find (a):[a^2 = b^2 + c^2 - 2bc cos A]Substituting (b = 2), (c = 4), and (cos 60^circ = frac{1}{2}):[a^2 = 2^2 + 4^2 - 2 times 2 times 4 times frac{1}{2}]Simplifying:[a^2 = 4 + 16 - 8 = 12 implies a = sqrt{12} = 2sqrt{3}]Thus, the value of (a) is (boxed{2sqrt{3}}).]

🔑:Given that in triangle ABC, A=60^{circ}, b=2, and S_{triangle ABC}=2 sqrt {3},We know that 2 sqrt {3}= frac {1}{2}bcsin A= frac {1}{2}times2times ctimes frac { sqrt {3}}{2}, solving for c we get c=4.Then, by using the cosine rule, we can find a:a^{2}=b^{2}+c^{2}-2bccos A=4+16-2times2times4times frac {1}{2}=12,Solving for a, we get: a=2 sqrt {3}.So, the answer is: boxed{2 sqrt {3}}.By using S_{triangle ABC}= frac {1}{2}bcsin A, we can find c, and then using the cosine rule, we can find a. This problem primarily tests the application of the formula for the area of a triangle and the cosine rule, and it is a basic problem.

🧠:[<think>Alright, let's try to solve this equation: (5 sqrt[15]{x^{22}} + sqrt[15]{x^{14} cdot sqrt{x}} - 22 sqrt[15]{x^{7}} = 0). Hmm, okay. First, I need to simplify those radicals. The radicals are all 15th roots, so maybe I can rewrite each term with exponents to make it easier.Starting with the first term: (5 sqrt[15]{x^{22}}). The 15th root of (x^{22}) is the same as (x^{22/15}). So, that term becomes (5x^{22/15}).Next term: (sqrt[15]{x^{14} cdot sqrt{x}}). Let's break that down. The (sqrt{x}) is (x^{1/2}), so inside the radical, we have (x^{14} cdot x^{1/2}). Adding the exponents, 14 + 1/2 = 14.5 or 29/2. So, inside the 15th root, it's (x^{29/2}). Then, taking the 15th root would be (x^{(29/2)/15} = x^{29/30}). So, the second term is (x^{29/30}).Third term: (-22 sqrt[15]{x^{7}}). Similarly, that's (-22x^{7/15}).So, rewriting the entire equation: (5x^{22/15} + x^{29/30} - 22x^{7/15} = 0). Hmm, let's see. The exponents here are 22/15, 29/30, and 7/15. To combine these terms, maybe we can express all exponents with a common denominator. The denominators are 15 and 30. Let's convert them all to 30.22/15 = 44/30, 29/30 stays the same, and 7/15 = 14/30. So, the equation becomes:(5x^{44/30} + x^{29/30} - 22x^{14/30} = 0).Hmm, noticing that 44/30 can be simplified as 22/15, which is the same as before, but maybe there's a substitution here. Let me think. If I let (y = x^{1/30}), then (x^{1/30} = y), so (x^{k/30} = y^k). Let's try that substitution.Let (y = x^{1/30}). Then:(x^{44/30} = y^{44}),(x^{29/30} = y^{29}),(x^{14/30} = y^{14}).So substituting into the equation:(5y^{44} + y^{29} - 22y^{14} = 0).Hmm, this seems like a polynomial equation in terms of y. But the exponents are still quite high. Let's see if we can factor out a common term. All terms have at least (y^{14}) as a factor. Let's factor that out:(y^{14}(5y^{30} + y^{15} - 22) = 0).So, setting each factor equal to zero:First factor: (y^{14} = 0) leads to (y = 0). But (y = x^{1/30}), so (x^{1/30} = 0) implies (x = 0). However, we need to check if x=0 is a valid solution. Plugging back into the original equation:First term: (5 sqrt[15]{0^{22}} = 0),Second term: (sqrt[15]{0^{14} cdot sqrt{0}} = 0),Third term: (-22 sqrt[15]{0^{7}} = 0). So, 0 + 0 - 0 = 0, which is true. So, x=0 is a solution.But usually, in equations with radicals, sometimes x=0 is considered extraneous, but here all terms are defined (since 15th roots are odd, so they can handle negative numbers, but x=0 is okay). So, x=0 is a valid solution.Second factor: (5y^{30} + y^{15} - 22 = 0). Let's let (z = y^{15}), so the equation becomes:(5z^2 + z - 22 = 0).Ah, that's a quadratic equation! Let's solve for z:Using quadratic formula: (z = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where a=5, b=1, c=-22.So:(z = frac{-1 pm sqrt{1 + 440}}{10}) because (b^2 - 4ac = 1 + 4*5*22 = 1 + 440 = 441).(sqrt{441} = 21), so:(z = frac{-1 pm 21}{10}).Thus, two solutions:1. (z = frac{-1 + 21}{10} = frac{20}{10} = 2).2. (z = frac{-1 - 21}{10} = frac{-22}{10} = -2.2).But remember that (z = y^{15}). Since y is defined as (x^{1/30}), which is a real number only if x is non-negative (since 30th root of a negative number isn't real). Wait, actually, 30 is even, so the 30th root of a negative number isn't real. Therefore, x must be non-negative. Therefore, y is real only if x is non-negative, so y is non-negative. Therefore, z = y^{15} is non-negative (since y is non-negative, and any power of non-negative is non-negative). Therefore, z cannot be negative, so z = -2.2 is invalid. So, only z = 2 is valid.So, z = 2 => (y^{15} = 2). Since y = x^{1/30}, then:((x^{1/30})^{15} = 2) => (x^{15/30} = 2) => (x^{1/2} = 2) => (sqrt{x} = 2) => x = 4.Therefore, the solutions are x=0 and x=4. But let's check them in the original equation to make sure.First, x=0: As before, all terms become zero, so 0 + 0 - 0 = 0. That's valid.Second, x=4:First term: (5 sqrt[15]{4^{22}}). Let's compute 4^22. 4 is 2^2, so 4^22 = (2^2)^22 = 2^44. Therefore, the 15th root of 2^44 is 2^(44/15). So, 5 * 2^(44/15).Second term: (sqrt[15]{4^{14} cdot sqrt{4}}). Compute 4^{14} = (2^2)^14 = 2^28. sqrt(4) = 2. So, 2^28 * 2 = 2^29. Then, the 15th root of 2^29 is 2^(29/15).Third term: -22 * sqrt[15]{4^7}. 4^7 = (2^2)^7 = 2^14. 15th root of 2^14 is 2^(14/15). So, -22 * 2^(14/15).So, the equation becomes:5*2^(44/15) + 2^(29/15) -22*2^(14/15) = 0.Hmm, let's factor out 2^(14/15):2^(14/15) * [5*2^(30/15) + 2^(15/15) -22] = 2^(14/15) * [5*2^2 + 2^1 -22] = 2^(14/15)*[5*4 + 2 -22] = 2^(14/15)*[20 + 2 -22] = 2^(14/15)*(22 -22) = 0.Yes, that works. So, x=4 is a valid solution.Therefore, the solutions are x=0 and x=4.But wait, let me check if there are more solutions. The substitution led us to y^{14}(5y^{30} + y^{15} -22)=0. So, y=0 gives x=0, and 5y^{30} + y^{15} -22=0 gives y^{15}=2. But since y = x^{1/30}, which is non-negative, are there other real solutions? When we solved for z, we had only z=2, which led to x=4. So, no other real solutions. Therefore, the only real solutions are x=0 and x=4.But let me check x=0 again. If x=0, then in the original equation, each term is:First term: 5 * 0^(22/15) = 5*0 =0Second term: 0^(14/15)*sqrt(0) inside the radical: 0^(14/15) is 0, sqrt(0)=0, so 0*0=0 inside the radical, which is 0. Then the 15th root of 0 is 0.Third term: -22*0^(7/15) = -22*0 =0.So, 0 + 0 -0=0. So, x=0 is valid.But sometimes, in equations with radicals, even if substituting x=0 gives 0=0, depending on the original problem's context, x=0 might be considered an extraneous solution. But in pure mathematical terms, since substituting x=0 doesn't cause any issues (no division by zero, no even roots of negative numbers), x=0 is a valid solution.Therefore, the solutions are x=0 and x=4. Let me see if the problem has any restrictions on x. The original equation has 15th roots and a square root. The square root requires that the argument is non-negative. So, sqrt(x) requires x >=0. Therefore, the domain of the equation is x >=0. Since x=0 and x=4 are both >=0, they are within the domain.Hence, both solutions are valid.Final AnswerThe solutions are (boxed{0}) and (boxed{4}).</think>Given the equation (5 sqrt[15]{x^{22}} + sqrt[15]{x^{14} cdot sqrt{x}} - 22 sqrt[15]{x^{7}} = 0), we start by rewriting each term with exponents.1. (5 sqrt[15]{x^{22}} = 5x^{22/15})2. (sqrt[15]{x^{14} cdot sqrt{x}} = sqrt[15]{x^{29/2}} = x^{29/30})3. (-22 sqrt[15]{x^{7}} = -22x^{7/15})Rewriting the equation with these exponents:[5x^{22/15} + x^{29/30} - 22x^{7/15} = 0]Next, we convert all exponents to a common denominator of 30:[5x^{44/30} + x^{29/30} - 22x^{14/30} = 0]We substitute (y = x^{1/30}), leading to:[5y^{44} + y^{29} - 22y^{14} = 0]Factoring out (y^{14}):[y^{14}(5y^{30} + y^{15} - 22) = 0]This gives two factors:1. (y^{14} = 0 implies y = 0 implies x = 0)2. (5y^{30} + y^{15} - 22 = 0)Letting (z = y^{15}), we solve the quadratic equation:[5z^2 + z - 22 = 0]Using the quadratic formula:[z = frac{-1 pm sqrt{1 + 440}}{10} = frac{-1 pm 21}{10}]This gives solutions:1. (z = 2)2. (z = -2.2) (invalid as (z) must be non-negative)For (z = 2):[y^{15} = 2 implies (x^{1/30})^{15} = 2 implies x^{1/2} = 2 implies x = 4]Checking both solutions in the original equation confirms they are valid. Therefore, the solutions are:[boxed{0} text{ and } boxed{4}],<think>Okay, so I need to solve this equation: (5 sqrt[15]{x^{22}} + sqrt[15]{x^{14} cdot sqrt{x}} - 22 sqrt[15]{x^{7}} = 0). Hmm, radicals can be tricky, especially with different exponents. Let me see how to simplify each term first. Starting with the first term: (5 sqrt[15]{x^{22}}). The 15th root of (x^{22}) can be written as (x^{22/15}). So that term becomes (5x^{22/15}). The second term is (sqrt[15]{x^{14} cdot sqrt{x}}). Let me break that down. The square root of x is (x^{1/2}), so inside the radical, we have (x^{14} cdot x^{1/2} = x^{14 + 1/2} = x^{29/2}). Then taking the 15th root of that gives (x^{29/(2 cdot 15)} = x^{29/30}). So the second term is (x^{29/30}). The third term is (-22 sqrt[15]{x^7}), which is (-22x^{7/15}). Putting it all together, the equation becomes:[5x^{22/15} + x^{29/30} - 22x^{7/15} = 0]Hmm, the exponents here are fractions with denominators 15, 30, and 15. Maybe I can find a common denominator or a substitution to simplify this. Let's see. The exponents are 22/15, 29/30, and 7/15. If I convert them all to 30 denominators:22/15 = 44/3029/30 stays the same.7/15 = 14/30So the equation is:[5x^{44/30} + x^{29/30} - 22x^{14/30} = 0]Looking at these exponents: 44/30, 29/30, 14/30. Hmm, 44/30 is 22/15, which is (7/15)*3 + 1/15. Not sure. Maybe factor out the smallest exponent? The smallest exponent here is 14/30 = 7/15. Let me try that.Let (y = x^{7/15}). Then, let's express each term in terms of y.First term: (5x^{44/30}). Since 44/30 = (44/30) = (22/15). But 22/15 can be written as (7/15)*3 + 1/15. Wait, 22 divided by 7 is 3 with a remainder of 1, but that might not help. Alternatively, 44/30 = 22/15 = (7/15 + 7/15 + 8/15). Not sure. Alternatively, since 22/15 = (7/15)*3 + 1/15. Maybe not helpful. Let me see.Wait, let me think. If y = x^{7/15}, then y^3 = x^{21/15} = x^{7/5}. But 44/30 is 22/15, which is 7/5 + 1/15. Hmm. Maybe not directly helpful. Alternatively, let's see:44/30 = (7/15)* (44/14) Hmm, 44 divided by 14 is 3.14... Not useful. Maybe another substitution.Alternatively, note that 44/30 = 22/15 = (14/15) + (8/15). Hmm, maybe not.Alternatively, express all exponents in terms of y = x^{1/30}, since the denominators are 30. Let's try that.Let y = x^{1/30}. Then:x^{44/30} = y^{44}x^{29/30} = y^{29}x^{14/30} = y^{14}So substituting into the equation:5y^{44} + y^{29} - 22y^{14} = 0Hmm, so now the equation is 5y^{44} + y^{29} - 22y^{14} = 0. That looks complicated because of the high exponents, but maybe factor out y^{14}?So factor out y^{14}:y^{14}(5y^{30} + y^{15} - 22) = 0Therefore, either y^{14} = 0 or 5y^{30} + y^{15} - 22 = 0But y = x^{1/30}, so y^{14} = 0 implies x = 0. However, plugging x = 0 into the original equation: first term is 5*0 = 0, second term is sqrt[15]{0} = 0, third term is -22*0 = 0. So 0 + 0 - 0 = 0. So x=0 is a solution. But is x=0 in the domain? The original equation has radicals with even roots in some terms? Wait, the original equation has a square root in the second term: sqrt(x). So sqrt(x) is defined for x >= 0. Then, all terms are defined for x >= 0. But if x=0, then each term is 0. So x=0 is a valid solution.But maybe the problem expects non-zero solutions? Let's check later.Now, the other factor is 5y^{30} + y^{15} - 22 = 0. Let's let z = y^{15}. Then, the equation becomes 5z^2 + z - 22 = 0.That's a quadratic equation! So solving for z:5z^2 + z - 22 = 0Using quadratic formula:z = [-1 ± sqrt(1 + 4*5*22)] / (2*5)Compute discriminant: 1 + 440 = 441sqrt(441) = 21Thus,z = [-1 ± 21]/10So two solutions:z = (-1 + 21)/10 = 20/10 = 2z = (-1 - 21)/10 = -22/10 = -11/5But z = y^{15} and y = x^{1/30}, so z = (x^{1/30})^{15} = x^{15/30} = x^{1/2} = sqrt(x). So sqrt(x) = 2 or sqrt(x) = -11/5But sqrt(x) is non-negative, so sqrt(x) = -11/5 is impossible. Thus, only sqrt(x) = 2. Therefore, x = 4.But we need to check if x=4 is a solution. Let's verify in the original equation.First term: 5 * 15th root of x^{22} = 5 * x^{22/15}x=4, so 5*4^{22/15}Second term: 15th root of (x^{14} * sqrt(x)) = (x^{14} * x^{1/2})^{1/15} = (x^{29/2})^{1/15} = x^{29/30}Third term: -22 * 15th root of x^7 = -22x^{7/15}So substituting x=4:First term: 5*4^{22/15}Second term: 4^{29/30}Third term: -22*4^{7/15}Let me compute 4^{1/15} first. Let’s denote k = 4^{1/15}. Then 4^{22/15} = k^{22}, 4^{29/30} = k^{29}, and 4^{7/15} = k^7.Then the equation becomes:5k^{22} + k^{29} - 22k^7 = 0But since k = 4^{1/15} is a positive real number. Let’s factor out k^7:k^7(5k^{15} + k^{22} - 22) = 0Wait, but k is 4^{1/15}, so k^{15} = 4^{1} = 4, and k^{22} = 4^{22/15} = 4^{1 + 7/15} = 4 * 4^{7/15} = 4k^7.Wait, this seems to complicate. Maybe there's a better way. Alternatively, since we found x=4 through substitution, maybe check if the exponents work out.Alternatively, notice that when we set z = y^{15} = sqrt(x), and found sqrt(x)=2, so x=4. Thus, the substitution leads us to x=4. Therefore, x=4 is a solution. Let me verify numerically.Compute each term with x=4:First term: 5 * (4)^{22/15}22/15 is approximately 1.4667, so 4^1.4667 ≈ 4^(1 + 0.4667) = 4 * 4^0.4667. Let's compute 4^0.4667. Since 4 = 2^2, so 4^0.4667 = 2^(2 * 0.4667) = 2^0.9334 ≈ 2^0.9334 ≈ 1.9 (since 2^1=2, 2^0.9≈1.866). Let's say approximately 1.9. So first term ≈ 5 * 4 * 1.9 ≈ 5 * 7.6 ≈ 38.Second term: (4)^{29/30}. 29/30 ≈ 0.9667. So 4^0.9667 ≈ 4^(1 - 0.0333) ≈ 4 / 4^0.0333. 4^0.0333 ≈ e^(ln4 * 0.0333) ≈ 1 + ln4 * 0.0333 (approx). ln4 ≈ 1.386, so 1.386 * 0.0333 ≈ 0.046. So approximately 1.046. Thus, 4 / 1.046 ≈ 3.825. So second term ≈ 3.825.Third term: -22 * 4^{7/15}. 7/15 ≈ 0.4667. 4^0.4667 ≈ same as before, approximately 1.9. So third term ≈ -22 * 1.9 ≈ -41.8.Thus total ≈ 38 + 3.825 - 41.8 ≈ 0.025. That's close to zero, considering the approximations. So x=4 is a valid solution.But let's check x=0 as well. When x=0, each term becomes zero, as we saw before. So x=0 is a solution.Are there any other solutions? Let's think. The quadratic equation gave us only z=2 and z=-11/5. The negative solution is rejected, so only z=2 leading to x=4. So the equation has two solutions: x=0 and x=4.But wait, we need to confirm if x=0 is indeed a solution. In the original equation:First term: 5 * 15th root of 0^22 = 0Second term: 15th root of (0^14 * sqrt(0)) = 15th root of 0 = 0Third term: -22 * 15th root of 0^7 = 0So 0 + 0 - 0 = 0, which is true. So x=0 is a valid solution.But sometimes, in equations with radicals, even if substituting x=0 yields 0, it might be considered an extraneous solution if, for example, some radicals are in denominators or if there are higher roots. But in this case, all radicals are 15th roots, which are defined for 0 (as 0). So x=0 is valid.Therefore, the solutions are x=0 and x=4.But wait, let me check if there are more solutions. The substitution process gave us x=0 and x=4. The substitution was y = x^{1/30}, leading to y^{14}(5y^{30} + y^{15} -22)=0. Then y=0 (x=0) or 5y^{30} + y^{15} -22=0. Then z = y^{15} leading to quadratic in z, which gave z=2 (x=4) and z=-11/5 (invalid). So only x=0 and x=4.But maybe in the process of substitution, we lost some solutions? Let's check. For example, when we set y = x^{1/30}, we assume x >=0 because radicals with even roots require non-negative x. So x must be >=0. So the original equation's domain is x >=0.Therefore, all possible solutions are x=0 and x=4.Wait, but let me check if x=1 satisfies the equation? Let's try x=1:First term: 5 * 15th root of 1^22 =5*1=5Second term: 15th root of (1^14 * sqrt(1))=15th root of 1=1Third term: -22*15th root of1^7=-22*1=-22So total:5 +1 -22= -16 ≠0. So x=1 is not a solution.x=4 and x=0 are the only solutions. Let me confirm with another test. Suppose x=16, would that do anything? Probably not. Let's check x=16.First term:5*(16)^{22/15}. Let's compute 16^{1/15}=2^{4/15}. So 16^{22/15}=2^{88/15}=2^5 *2^{13/15}≈32 *1.7≈54.4. So first term≈5*54.4≈272Second term:16^{29/30}=2^{4*29/30}=2^{116/30}=2^{58/15}=2^3*2^(13/15)≈8*1.7≈13.6Third term:-22*16^{7/15}= -22*(2^{28/15})=-22*(2^1 *2^{13/15})≈-22*2*1.7≈-22*3.4≈-74.8Total≈272 +13.6 -74.8≈210.8, which is not zero. So x=16 is not a solution.Therefore, only x=0 and x=4 are solutions. Wait, but x=0, when plugged into the original equation, gives all terms zero, which works. However, sometimes in equations, even if substitution gives 0, but if the equation is derived from some other steps, x=0 might not be acceptable. But in this case, since all radicals are defined at x=0, and substituting gives a valid equation, it's a legitimate solution.Therefore, the solutions are x=0 and x=4.But let me check the problem statement again. It says "put your final answer within boxed{}". It might be expecting multiple answers. So maybe the answer is x=0 and x=4. But let me check once again.Original equation:(5 sqrt[15]{x^{22}} + sqrt[15]{x^{14} cdot sqrt{x}} - 22 sqrt[15]{x^{7}} = 0)We found x=0 and x=4. Let me check x=4 again more precisely.First term: 5*(4)^{22/15}4 is 2^2, so 4^{22/15} = (2^2)^{22/15}=2^{44/15}=2^(2 + 14/15)=4 * 2^(14/15). 2^(14/15) is approximately 2^0.933≈1.9. So approx 4*1.9≈7.6. Multiply by 5:≈38.Second term: sqrt[15]{4^{14} * sqrt(4)}. sqrt(4)=2. 4^{14}=2^{28}. So inside the radical:2^{28}*2=2^{29}. 15th root of 2^{29}=2^{29/15}=2^(1 +14/15)=2*2^(14/15)≈2*1.9≈3.8.Third term:-22*sqrt[15]{4^7}= -22*(4^{7/15})= -22*(2^{14/15})≈-22*1.9≈-41.8.Total≈38 +3.8 -41.8≈0. So exact value would be zero. Therefore, x=4 is exact.So the solutions are x=0 and x=4.But the problem might be expecting only x=4, maybe considering x=0 trivial? But since it's a valid solution, we should include both. Wait, maybe in the original equation, x=0 makes all terms zero, but perhaps the problem is designed for x>0? Not sure. But according to the mathematics, x=0 and x=4 are both solutions.Wait, but let me check with the substitution steps. When we set y = x^{1/30}, y=0 corresponds to x=0. Then, the equation y^{14}(5y^{30} + y^{15} -22)=0 gives y=0 or 5y^{30} + y^{15} -22=0. So y=0 (x=0) and solutions from the quadratic. Therefore, both x=0 and x=4 are valid.Therefore, the solutions are x=0 and x=4. So I need to present both. The problem didn't specify any constraints, so both are valid.But let me check once again for x=0. Original equation:First term:5 * sqrt[15]{0^{22}} =5*0=0Second term:sqrt[15]{0^{14}*sqrt{0}} = sqrt[15]{0}=0Third term:-22*sqrt[15]{0^7}= -22*0=0Thus, 0 +0 -0=0. Correct.Therefore, both x=0 and x=4 are solutions. However, sometimes problems of this nature are posed in contexts where x=0 is considered trivial or not of interest, but unless specified, both are valid.Hence, the solutions are x=0 and x=4. So I should box both.Final AnswerThe solutions are boxed{0} and boxed{4}.</think>To solve the equation (5 sqrt[15]{x^{22}} + sqrt[15]{x^{14} cdot sqrt{x}} - 22 sqrt[15]{x^{7}} = 0), we start by simplifying each term using fractional exponents:1. The first term (5 sqrt[15]{x^{22}}) simplifies to (5x^{22/15}).2. The second term (sqrt[15]{x^{14} cdot sqrt{x}}) simplifies to (x^{29/30}).3. The third term (-22 sqrt[15]{x^7}) simplifies to (-22x^{7/15}).Converting all exponents to a common denominator of 30, we get:[5x^{44/30} + x^{29/30} - 22x^{14/30} = 0]Next, we factor out the smallest exponent (x^{14/30}):[x^{14/30} left(5x^{30/30} + x^{15/30} - 22right) = 0]Let (y = x^{1/30}). Substituting (y) into the equation, we get:[y^{14} left(5y^{30} + y^{15} - 22right) = 0]This gives us two cases:1. (y^{14} = 0) which implies (y = 0) and thus (x = 0).2. Solving the quadratic equation (5y^{30} + y^{15} - 22 = 0) by letting (z = y^{15}), we get: [ 5z^2 + z - 22 = 0 ] Using the quadratic formula, we find (z = 2) (since the negative solution is discarded). This gives (y^{15} = 2) and thus (x = 4).Verifying both solutions in the original equation:- For (x = 0), all terms are zero, confirming it is a solution.- For (x = 4), substituting back confirms the equation holds true.Thus, the solutions are (boxed{0}) and (boxed{4}).]

🔑:## Solution.Domain of definition: x geq 0.From the condition 5 x^{frac{22}{15}}+x^{frac{27}{30}}-22 x^{frac{7}{15}}=0 Leftrightarrow 5 x^{frac{44}{30}}+x^{frac{27}{30}}-22 x^{frac{14}{30}}=0 Leftrightarrow Leftrightarrow x^{frac{14}{30}}left(5 x+x^{frac{1}{2}}-22right)=0. Then x^{frac{14}{30}}=0, x_{1}=0 or 5 x+x^{frac{1}{2}}-22=0.Let x^{frac{1}{2}}=y geq 0. The equation in terms of y becomes5 y^{2}+y-22=0, from which y_{1}=-frac{11}{5}, y_{2}=2 ; y_{1}=-frac{11}{5}<0 is not valid. Then x_{2}^{frac{1}{2}}=2, x_{2}=4Answer: x_{1}=0, x_{2}=4

🧠:[<think>Okay, let's tackle this problem step by step. The question is asking which operation principle isn't used when calculating 2√3 multiplied by 5√6 to get 30√2. The options are Distributive property, Commutative/Associative properties, Multiplication rule of square roots, and the Property √a² = a for a ≥ 0.First, I'll recall how to multiply expressions with square roots. When multiplying two radicals, you can multiply the coefficients together and then the radicands (the numbers under the roots) together. So, for 2√3 * 5√6, I would multiply 2 and 5 first, which gives 10. Then multiply √3 and √6. Wait, the multiplication rule of square roots is √a * √b = √(ab), right? So that would be √(3*6) = √18. Then √18 can be simplified because 18 is 9*2, and √9 is 3. So √18 = 3√2. Then multiplying that by the 10 from earlier gives 10*3√2 = 30√2. Now, let's look at each step and see which properties were used. First, multiplying the coefficients 2 and 5: that's straightforward multiplication, which uses the commutative and associative properties of multiplication, right? Because when you rearrange the order (commutative) or group them differently (associative), the result remains the same. So 2*5 is 5*2, and that's commutative. So that would relate to option B.Then, multiplying √3 and √6 using the multiplication rule of square roots (option C). That's definitely used here, as per the rule √a * √b = √(ab). So that's option C.Next, simplifying √18 into 3√2. That step uses the property of square roots where you can break down the radicand into factors, specifically using √(a²*b) = a√b. Here, 18 is 9*2, and 9 is 3². So √(9*2) = √9 * √2 = 3√2. That seems like using the property that √(a²) = a when a ≥ 0. So that's option D. Because √9 = √(3²) = 3. So option D is used here.Now, the question is which property is NOT used. The options given are A, B, C, D. So far, we've used B, C, D. What about the distributive property (option A)? The distributive property is about distributing multiplication over addition, like a(b + c) = ab + ac. In this problem, we aren't adding anything; it's just multiplication of terms. There's no addition involved here, so the distributive property isn't used. Let me double-check. The original expression is 2√3 * 5√6. To compute this, we rearrange the multiplication as (2*5)*(√3*√6). This rearrangement is allowed due to the commutative and associative properties of multiplication (changing the order and grouping), which is option B. Then applying the multiplication rule for square roots (option C) gives √18, which is simplified using the property of square roots (option D). Since there's no addition or subtraction involved, the distributive property, which deals with distributing over addition, isn't used at all. Therefore, the answer should be option A.Final Answerboxed{A}</think>To calculate the result of (2sqrt{3} times 5sqrt{6}), we follow these steps:1. Multiply the coefficients: (2 times 5 = 10). This step uses the commutative and associative properties of multiplication (option B).2. Multiply the radicands: (sqrt{3} times sqrt{6} = sqrt{18}). This step uses the multiplication rule of square roots (option C).3. Simplify (sqrt{18}) as (sqrt{9 times 2} = sqrt{9} times sqrt{2} = 3sqrt{2}). This step uses the property of square roots (sqrt{a^2} = a) for (a geq 0) (option D).4. Combine the results: (10 times 3sqrt{2} = 30sqrt{2}).The distributive property (option A) is not used in any step of this calculation because there is no addition or subtraction involved. (boxed{A}),<think>Okay, let me try to figure out this problem. So, the question is asking which operation principle isn't used when calculating 2√3 multiplied by 5√6, resulting in 30√2. The options are the distributive property, commutative/associative properties, multiplication rule of square roots, and the property that √a² = a for a ≥ 0.First, let me go through the calculation step by step. The original expression is 2√3 × 5√6. To solve this, I think you would first multiply the coefficients (the numbers outside the square roots) and then multiply the radicals (the square roots). So, multiplying 2 and 5 gives 10. Then, multiplying √3 and √6. Wait, when multiplying square roots, there's a rule that √a × √b = √(a×b). That must be the multiplication rule of square roots, which is option C. So, √3 × √6 becomes √(3×6) = √18. Then, √18 can be simplified because 18 is 9×2, and √9 is 3. So, √18 simplifies to 3√2. Therefore, putting it all together, 10 × 3√2 = 30√2. Now, let's analyze which properties were used here. First, when we multiplied 2 and 5, and √3 and √6 separately, did we use the commutative or associative properties? The commutative property says that the order of multiplication doesn't matter, and associative is about grouping. Here, when we rearrange the multiplication from 2×√3×5×√6 to (2×5)×(√3×√6), we're using commutative (switching the order) and associative (grouping differently) properties. So that would be option B.Then, using the multiplication rule for square roots (option C) as I mentioned earlier. Then, simplifying √18 to 3√2 uses the property that √(a²×b) = a√b, right? So √18 is √(9×2) = √9×√2 = 3√2. Here, √9 is 3 because 9 is 3 squared. That relates to the property √a² = a when a is non-negative, which is option D. So we did use that when simplifying √9 to 3.Now, what about the distributive property (option A)? The distributive property involves multiplication over addition or subtraction, like a(b + c) = ab + ac. But in this problem, there's no addition or subtraction involved. We're just multiplying terms together. So the distributive property wasn't used here. Therefore, the answer should be option A.Wait, let me double-check. The original problem is a straightforward multiplication, so commutative and associative allow us to rearrange and group the terms, multiplication rule of roots combines the radicals, and then the square root of a square simplifies. But no distribution because there's no sum or difference to distribute over. Yeah, so the one not used is the distributive property. So the answer is A.Final Answerboxed{A}</think>To calculate the result of (2sqrt{3} times 5sqrt{6}), we follow these steps:1. Multiply the coefficients: (2 times 5 = 10).2. Multiply the radicals: (sqrt{3} times sqrt{6} = sqrt{3 times 6} = sqrt{18}). This uses the multiplication rule of square roots (option C).3. Simplify (sqrt{18}): (sqrt{18} = sqrt{9 times 2} = sqrt{9} times sqrt{2} = 3sqrt{2}). This uses the property (sqrt{a^2} = a) (for (a geq 0)) (option D).4. Combine the results: (10 times 3sqrt{2} = 30sqrt{2}).During this process, the commutative and associative properties of multiplication (option B) are used to rearrange and group the terms. The distributive property (option A) is not used because there is no addition or subtraction involved in the expression.Thus, the operation principle that is not used is the distributive property.(boxed{A})]

🔑:To calculate 2sqrt{3} times 5sqrt{6}, we follow these steps:1. Apply the Commutative and Associative Properties of Multiplication:We first rearrange and group the terms to simplify the calculation.[2sqrt{3} times 5sqrt{6} = 2 times 5 times (sqrt{3} times sqrt{6})]2. Simplify the Multiplication:Next, we multiply the numbers outside the square roots and then apply the multiplication rule of square roots.[= 10 times sqrt{3 times 6}]3. Multiply Inside the Square Root:We find the product inside the square root.[= 10sqrt{18}]4. Simplify the Square Root:Recognizing that 18 can be expressed as 9 times 2 or {3}^{2} times 2, we simplify the square root.[= 10sqrt{{3}^{2} times 2}]5. Apply the Property of Square Roots sqrt{a^2}=a(a≥0):We simplify the square root of a square number.[= 10 times 3sqrt{2}]6. Final Calculation:Finally, we multiply the numbers outside the square root.[= 30sqrt{2}]In the calculation process, we utilized the commutative and associative properties of multiplication, the multiplication rule of square roots, and the property of square roots sqrt{a^2}=a(a≥0). The distributive property of multiplication was not used in this process.Therefore, the correct answer is:[boxed{text{A}}]

🧠:[<think>Okay, so I need to figure out the fifth inequality following the given pattern. Let me start by looking at the examples provided:1. (1 + frac{1}{2^{2}} < frac{3}{2})2. (1 + frac{1}{2^{2}} + frac{1}{3^{2}} < frac{5}{3})3. (1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} < frac{7}{4})Hmm, so each inequality adds the reciprocal square of the next integer. The first one has terms up to (2^2), the second up to (3^2), the third up to (4^2), so the pattern is that the nth inequality includes terms from (1) to (frac{1}{(n+1)^2}). Let me verify:First inequality (n=1?): terms 1 and (1/2^2), right-hand side (RHS) is (3/2). Wait, maybe the index is different. Let's see:First inequality: terms up to (1/2^2), RHS (3/2 = (2*1 +1)/2). Second inequality: terms up to (1/3^2), RHS (5/3 = (2*2 +1)/3). Third inequality: terms up to (1/4^2), RHS (7/4 = (2*3 +1)/4). So the pattern for the RHS seems to be when adding up to (1/(k)^2), the RHS is (frac{2(k-1) + 1}{k}). Wait, let's check:First inequality: up to (1/2^2), so k=2. RHS is (2*(2-1) +1)/2 = (2*1 +1)/2 = 3/2. Correct.Second inequality: up to (1/3^2), k=3. RHS is (2*(3-1) +1)/3 = (4 +1)/3 = 5/3. Correct.Third inequality: up to (1/4^2), k=4. RHS is (2*(4-1) +1)/4 = (6 +1)/4 = 7/4. Correct.Therefore, generalizing, the nth inequality is the sum from i=1 to k-1 of (1/i^2) where k = n + 1? Wait, let me check. Wait, first inequality is k=2, which would be n=1. Second inequality is k=3 (n=2). Third inequality is k=4 (n=3). So the nth inequality is sum from i=1 to k-1 where k = n +1? Wait, no, the first inequality is sum up to 1/2^2 (i=2), but written as 1 + 1/2^2. So actually, the number of terms is n +1. Wait, first inequality has two terms: 1 and 1/2^2. Second inequality has three terms: 1, 1/2^2, 1/3^2. Third inequality has four terms: 1, 1/2^2, 1/3^2, 1/4^2. So the nth inequality has n+1 terms. Therefore, the fifth inequality should have 5 +1 =6 terms, right? So terms from 1 to 1/6^2. Wait, but the first inequality is n=1, with terms up to 1/2^2. So maybe the term count is n +1. So for the fifth inequality, n=5, terms up to 1/(5+1)^2 =1/6^2. Then the sum would be 1 + 1/2^2 + 1/3^2 + 1/4^2 +1/5^2 +1/6^2. Then the RHS would follow the pattern: for the kth term, where k = n +1 (since first inequality is k=2), then RHS is (2(k -1) +1)/k. Wait, let me check with previous examples.Wait, for k=2 (n=1), RHS is (2*(2 -1)+1)/2 =3/2. Correct. For k=3 (n=2), RHS is (2*(3 -1)+1)/3=5/3. Correct. For k=4 (n=3), RHS is (2*(4 -1)+1)/4=7/4. So if the fifth inequality is n=5, then k=6. Then RHS would be (2*(6 -1) +1)/6 = (10 +1)/6=11/6. So the fifth inequality is:1 + 1/2² + 1/3² + 1/4² +1/5² +1/6² < 11/6.But let me verify if this pattern holds. Let me check the existing examples:First inequality: sum up to 1/2²: 1 + 1/4 = 5/4 ≈1.25. RHS is 3/2=1.5. So 1.25 <1.5: correct.Second inequality: sum up to 1/3²: 1 +1/4 +1/9 ≈1 +0.25 +0.111≈1.361. RHS is5/3≈1.666. Correct.Third inequality: sum up to1/4²:≈1 +0.25 +0.111 +0.0625≈1.4235. RHS 7/4=1.75. Correct.Continuing, the fourth inequality (if there was one listed) would be sum up to1/5², and RHS would be (2*(5-1)+1)/5=(8 +1)/5=9/5=1.8. Let's compute the sum: previous sum≈1.4235 +1/25≈1.4235 +0.04≈1.4635 <1.8. Correct.Then fifth inequality would be sum up to1/6²≈1.4635 +1/36≈1.4635 +0.0277≈1.4912. RHS=11/6≈1.8333. So 1.4912 <1.8333. Correct. So the pattern holds.Therefore, the fifth inequality is:1 + 1/2² +1/3² +1/4² +1/5² +1/6² <11/6.So the answer should be the sum up to 1/6² on the left and 11/6 on the right. But let me check how the question is phrased: "Following this pattern, the fifth inequality is ______." So they probably want the fifth inequality written out with the sum and the fraction. Let me check the initial examples:First inequality: terms 1 +1/2² <3/2Second:1 +1/2² +1/3² <5/3Third:1 +1/2² +1/3² +1/4² <7/4So the pattern is each next inequality adds the next reciprocal square. So the fourth would add 1/5², fifth adds 1/6². Therefore, the fifth inequality is:1 + 1/2² +1/3² +1/4² +1/5² +1/6² <11/6.Therefore, written in LaTeX, it should be:1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} + frac{1}{5^{2}} + frac{1}{6^{2}} < frac{11}{6}But let me verify once again. Since the first three inequalities are given as:1. 1 +1/2² <3/2 (terms up to 2)2. 1 +1/2² +1/3² <5/3 (terms up to3)3. 1 +1/2² +1/3² +1/4² <7/4 (terms up to4)Hence, the nth inequality has terms up to 1/(n+1)². Therefore, the fifth inequality would have terms up to 1/(5+1)² =1/6². Hence, sum from 1 to1/6², and RHS is (2*(6-1)+1)/6=11/6. Correct.Therefore, the fifth inequality is as above.Final AnswerThe fifth inequality is boxed{1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} + frac{1}{5^{2}} + frac{1}{6^{2}} < frac{11}{6}}.</think>The given inequalities follow a pattern where each inequality adds the reciprocal square of the next integer. The right-hand side (RHS) of each inequality follows the formula (frac{2(k-1) + 1}{k}) where (k) is the next integer after the last term in the left-hand side (LHS).1. First inequality: (1 + frac{1}{2^{2}} < frac{3}{2})2. Second inequality: (1 + frac{1}{2^{2}} + frac{1}{3^{2}} < frac{5}{3})3. Third inequality: (1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} < frac{7}{4})For each inequality, the RHS is calculated as follows:- For (k = 2): (frac{2(2-1) + 1}{2} = frac{3}{2})- For (k = 3): (frac{2(3-1) + 1}{3} = frac{5}{3})- For (k = 4): (frac{2(4-1) + 1}{4} = frac{7}{4})Following this pattern, the fifth inequality will include terms up to (frac{1}{6^{2}}) (since the fifth inequality corresponds to (k = 6)). The RHS for (k = 6) is:[frac{2(6-1) + 1}{6} = frac{11}{6}]Thus, the fifth inequality is:[1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} + frac{1}{5^{2}} + frac{1}{6^{2}} < frac{11}{6}][boxed{1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} + frac{1}{5^{2}} + frac{1}{6^{2}} < frac{11}{6}}],<think>Okay, let's try to figure out the fifth inequality in the given pattern. The problem shows the first three inequalities:1. (1 + frac{1}{2^{2}} < frac{3}{2})2. (1 + frac{1}{2^{2}} + frac{1}{3^{2}} < frac{5}{3})3. (1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} < frac{7}{4})And it mentions that following this pattern, the fifth inequality is _____. So, I need to identify the pattern in both the left-hand side (LHS) and the right-hand side (RHS) of the inequalities to determine what the fifth one should be.First, let's analyze the left-hand side. Each inequality adds another term to the sum. The first inequality has two terms: 1 and ( frac{1}{2^2} ). The second has three terms, adding ( frac{1}{3^2} ), and the third has four terms, adding ( frac{1}{4^2} ). So, each subsequent inequality includes one more reciprocal square term. Therefore, the pattern for the LHS is:For the nth inequality, the LHS is ( 1 + frac{1}{2^2} + frac{1}{3^2} + dots + frac{1}{(n+1)^2} ).Wait, let's check that. The first inequality is the fifth one? Wait, no. Let's see. The first inequality given is the first one: 1 + 1/2² < 3/2. Then the second inequality adds 1/3², so it's up to 1/3². The third adds 1/4². So, each inequality adds one more term. Therefore, the first inequality has terms up to 1/2², the second up to 1/3², the third up to 1/4², so the nth inequality has terms up to 1/(n+1)². So, if we need the fifth inequality, it should have terms up to 1/(5+1)² = 1/6². So, the LHS would be 1 + 1/2² + 1/3² + 1/4² + 1/5² + 1/6².Wait, but let's confirm with the given examples. The first inequality is 1 + 1/2² < 3/2. So, that's two terms: 1 and 1/2², which is up to n=2. The next one adds 1/3², so n=3, etc. So the pattern is that the kth inequality has terms from 1 up to 1/(k+1)². Therefore, the fifth inequality will have terms from 1 up to 1/(5+1)² = 1/6². So the LHS for the fifth inequality is:(1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + frac{1}{6^2}).Now, the RHS of the inequalities are 3/2, 5/3, 7/4. Let's see if there's a pattern here. Let's note the denominators and numerators:First inequality RHS: 3/2 → numerator 3, denominator 2.Second inequality RHS: 5/3 → numerator 5, denominator 3.Third inequality RHS: 7/4 → numerator 7, denominator 4.Looking at the sequence of numerators: 3, 5, 7. That's an arithmetic sequence increasing by 2 each time. Similarly, denominators: 2, 3, 4, increasing by 1 each time.Therefore, the pattern for the RHS is numerator = 2k + 1 and denominator = k + 1, where k is the inequality number starting from 1.Let's check:For k=1: numerator = 2*1 + 1 = 3, denominator = 1 + 1 = 2 → 3/2. Correct.For k=2: numerator = 2*2 + 1 = 5, denominator = 2 + 1 = 3 → 5/3. Correct.For k=3: numerator = 2*3 + 1 = 7, denominator = 3 + 1 = 4 → 7/4. Correct.Therefore, for the fifth inequality (k=5), the RHS should be numerator = 2*5 + 1 = 11, denominator = 5 + 1 = 6. So the RHS is 11/6.Therefore, the fifth inequality should be:(1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + frac{1}{6^2} < frac{11}{6}).But let me double-check this pattern. Let's see if this continues. Let's check the fourth inequality, which isn't given but perhaps if we can infer it.Following the pattern, the fourth inequality's LHS would add 1/5² to the previous one (third inequality had up to 1/4², so fourth would add 1/5²). Therefore, the fourth inequality would be:(1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} < frac{9}{5}).Since numerator for k=4: 2*4 +1 =9, denominator=5. So, 9/5. Then the fifth inequality, as above, adds 1/6² and RHS is 11/6.But maybe we should verify if this is indeed true. Let's compute the actual sums and check if the inequalities hold.First inequality: 1 + 1/4 = 1.25 < 3/2 = 1.5. True.Second inequality: 1 + 1/4 + 1/9 ≈ 1 + 0.25 + 0.1111 ≈ 1.3611 < 5/3 ≈ 1.6667. True.Third inequality: 1 + 1/4 + 1/9 + 1/16 ≈ 1 + 0.25 + 0.1111 + 0.0625 ≈ 1.4236 < 7/4 = 1.75. True.Fourth inequality (predicted): 1 + 1/4 + 1/9 + 1/16 + 1/25 ≈ 1 + 0.25 + 0.1111 + 0.0625 + 0.04 ≈ 1.4636 < 9/5 = 1.8. True.Fifth inequality: sum up to 1/6². Let's compute that.1 (term 1) + 1/4 (≈0.25) + 1/9 (≈0.1111) + 1/16 (≈0.0625) + 1/25 (≈0.04) + 1/36 (≈0.0278). Adding these:1 + 0.25 = 1.25+ 0.1111 ≈ 1.3611+ 0.0625 ≈ 1.4236+ 0.04 ≈ 1.4636+ 0.0278 ≈ 1.4914So the sum is approximately 1.4914. The RHS is 11/6 ≈ 1.8333. So indeed, 1.4914 < 1.8333. So the inequality holds.Therefore, the pattern seems consistent. Each time, we add the next reciprocal square term to the LHS, and the RHS follows the pattern (2k +1)/(k +1) where k is the inequality number.Therefore, the fifth inequality is:(1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} + frac{1}{5^{2}} + frac{1}{6^{2}} < frac{11}{6}).To confirm, let's write out the progression:1st inequality: k=1, terms up to 1/2², RHS=3/2.2nd inequality: k=2, terms up to 1/3², RHS=5/3.3rd inequality: k=3, terms up to 1/4², RHS=7/4.4th inequality: k=4, terms up to 1/5², RHS=9/5.5th inequality: k=5, terms up to 1/6², RHS=11/6.Yes, that seems correct. Each time, the RHS is (2k+1)/(k+1). So for the fifth inequality, k=5, so (2*5 +1)/(5 +1)=11/6.Therefore, the answer should be the sum up to 1/6² on the left and 11/6 on the right. So the fifth inequality is:(1 + frac{1}{2^{2}} + frac{1}{3^{2}} + frac{1}{4^{2}} + frac{1}{5^{2}} + frac{1}{6^{2}} < frac{11}{6}).Hence, that's the answer.Final AnswerThe fifth inequality is boxed{1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + frac{1}{6^2} < frac{11}{6}}.</think>The given inequalities follow a pattern where each subsequent inequality adds one more reciprocal square term to the left-hand side (LHS) and the right-hand side (RHS) follows a specific sequence. 1. The first inequality is (1 + frac{1}{2^2} < frac{3}{2}).2. The second inequality is (1 + frac{1}{2^2} + frac{1}{3^2} < frac{5}{3}).3. The third inequality is (1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} < frac{7}{4}).The pattern for the LHS is that each inequality adds the next reciprocal square term. For the fifth inequality, the LHS will include terms up to (frac{1}{6^2}).The RHS follows the pattern (frac{2k + 1}{k + 1}) where (k) is the inequality number. For the fifth inequality ((k = 5)), the RHS is (frac{11}{6}).Thus, the fifth inequality is:[1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + frac{1}{6^2} < frac{11}{6}][boxed{1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + frac{1}{6^2} < frac{11}{6}}]]

🔑:From the given inequalities: 1+ frac {1}{2^{2}} < frac {3}{2}, 1+ frac {1}{2^{2}}+ frac {1}{3^{2}} < frac {5}{3}, 1+ frac {1}{2^{2}}+ frac {1}{3^{2}}+ frac {1}{4^{2}} < frac {7}{4}, (ldots) We can induce that the nth inequality is: 1+ frac {1}{2^{2}}+ frac {1}{3^{2}}+ldots+ frac {1}{(n+1)^{2}} < frac {2n+1}{n+1}, When n=5, the fifth inequality is 1+ frac {1}{2^{2}}+ frac {1}{3^{2}}+ frac {1}{4^{2}}+ frac {1}{5^{2}}+ frac {1}{6^{2}} < frac {11}{6}, Therefore, the answer is: 1+ frac {1}{2^{2}}+ frac {1}{3^{2}}+ frac {1}{4^{2}}+ frac {1}{5^{2}}+ frac {1}{6^{2}} < frac {11}{6} By analyzing the pattern of change on both sides of the given inequalities 1+ frac {1}{2^{2}} < frac {3}{2}, 1+ frac {1}{2^{2}}+ frac {1}{3^{2}} < frac {5}{3}, 1+ frac {1}{2^{2}}+ frac {1}{3^{2}}+ frac {1}{4^{2}} < frac {7}{4}, (ldots), we can find the answer. This problem tests inductive reasoning, which generally involves: (1) observing individual cases to discover some common properties, and (2) deducing a general proposition from these known common properties.Thus, the answer is boxed{1+ frac {1}{2^{2}}+ frac {1}{3^{2}}+ frac {1}{4^{2}}+ frac {1}{5^{2}}+ frac {1}{6^{2}} < frac {11}{6}}.

🧠:[<think>Okay, so the problem is about two six-sided dice. Each die is fair in that each face is equally likely, but one die has a 4 replaced by a 3, and the other die has a 3 replaced by a 4. We need to find the probability that the sum of the numbers rolled on these two dice is odd. The answer choices are given from A to E.First, let me make sure I understand the setup correctly. Each die is still six-sided and fair, meaning each face has a 1/6 chance. The first die has the 4 replaced by 3. So, let's call that Die A. The original faces of a die are 1, 2, 3, 4, 5, 6. So Die A would have 1, 2, 3, 3, 5, 6. The second die, Die B, has the 3 replaced by 4. So its faces would be 1, 2, 4, 4, 5, 6. Is that right? Let me check. Original die: 1, 2, 3, 4, 5, 6. Die A: replace 4 with 3, so two 3s. Die B: replace 3 with 4, so two 4s. Yes, that seems correct.So, Die A has numbers: 1, 2, 3, 3, 5, 6. Die B has numbers: 1, 2, 4, 4, 5, 6. Each die is rolled once, and we want the probability that their sum is odd.Hmm. Okay, probability of odd sum. Remember that the sum of two numbers is odd if one is even and the other is odd. If both are even or both are odd, their sum is even. So the probability that the sum is odd is equal to the probability that one die is even and the other is odd.So, perhaps we can compute the probability of Die A being even and Die B being odd, plus the probability of Die A being odd and Die B being even.So, to do that, first we need to know the probability that Die A is even, and the probability that Die A is odd. Similarly for Die B.Let's analyze Die A first. Die A has faces: 1, 2, 3, 3, 5, 6.Numbers on Die A: 1 (odd), 2 (even), 3 (odd), 3 (odd), 5 (odd), 6 (even). So, how many odd numbers and even numbers are there?Odd numbers: 1, 3, 3, 5. That's four numbers. Even numbers: 2, 6. That's two numbers. Therefore, the probability Die A shows an odd number is 4/6 = 2/3, and even is 2/6 = 1/3.Similarly, Die B has faces: 1, 2, 4, 4, 5, 6.Numbers on Die B: 1 (odd), 2 (even), 4 (even), 4 (even), 5 (odd), 6 (even). So odd numbers: 1, 5. Two numbers. Even numbers: 2, 4, 4, 6. That's four numbers. Therefore, probability Die B is odd is 2/6 = 1/3, and even is 4/6 = 2/3.Okay, so Die A has a 2/3 chance of odd, Die B has a 1/3 chance of odd. So, the probability that Die A is odd and Die B is even is (2/3)*(2/3) = 4/9. The probability that Die A is even and Die B is odd is (1/3)*(1/3) = 1/9. Then total probability is 4/9 + 1/9 = 5/9. Hmm, that's answer choice D.Wait, but let me double-check. Maybe I made a mistake in counting the odd and even numbers on each die. Let's recheck.For Die A: original numbers 1, 2, 3, 4, 5, 6. But 4 is replaced by 3, so numbers become 1, 2, 3, 3, 5, 6. So yes, odd numbers are 1, 3, 3, 5: four odds. Even numbers are 2, 6: two evens. So 4/6 and 2/6. Correct.For Die B: original numbers 1, 2, 3, 4, 5, 6. But 3 is replaced by 4, so numbers become 1, 2, 4, 4, 5, 6. So odd numbers are 1, 5: two odds. Even numbers are 2, 4, 4, 6: four evens. So 2/6 and 4/6. Correct.So, the probability that Die A is odd and Die B is even: Die A odd is 4/6, Die B even is 4/6. So 4/6 * 4/6 = 16/36 = 4/9. Wait, no, hold on. Wait, Die B even is 4/6, so Die A odd is 4/6, Die B even is 4/6. So 4/6 * 4/6 = 16/36. Then Die A even (2/6) * Die B odd (2/6) = 4/36. So total is 16/36 + 4/36 = 20/36 = 5/9. Yes, that's 5/9, which is option D. So that seems to check out.But let me think if there's another way to approach this. Maybe list all possible outcomes? But since each die has 6 sides, there are 6*6=36 possible outcomes. But that might take a while, but maybe for verification.Alternatively, think about the parity of each die. Let's consider Die A and Die B.For Die A: numbers are 1,2,3,3,5,6. Let's list the parities:1: odd2: even3: odd3: odd5: odd6: evenSo, as before, four odds, two evens.For Die B: numbers are 1,2,4,4,5,6. Parities:1: odd2: even4: even4: even5: odd6: evenSo two odds, four evens.So, the possible combinations: (Odd, Odd), (Odd, Even), (Even, Odd), (Even, Even).The sum is odd only when (Odd, Even) or (Even, Odd). So the total probability is P(A odd)*P(B even) + P(A even)*P(B odd) = (4/6)(4/6) + (2/6)(2/6) = (16 + 4)/36 = 20/36 = 5/9. So same answer.Alternatively, if I list all possible outcomes, but that might be tedious. But perhaps we can check with an example. Let's see.First, Die A's numbers: 1,2,3,3,5,6. Die B's numbers:1,2,4,4,5,6.So for each number on Die A, paired with each number on Die B, check if the sum is odd. Then count how many of the 36 pairs result in an odd sum.But since it's 36 pairs, maybe we can compute how many odd sums there are.Alternatively, note that for each die, the number of odd numbers on Die A is 4, even is 2. For Die B, number of odd numbers is 2, even is 4.To get an odd sum, you need one odd and one even. So the number of such pairs is (number of odd on A)*(number of even on B) + (number of even on A)*(number of odd on B).Which is 4*4 + 2*2 = 16 + 4 = 20. So 20 out of 36, which is 5/9. Yep, same answer.Therefore, the answer should be D) 5/9.But just to be thorough, let me check with a specific enumeration.Take Die A's odd numbers: 1,3,3,5 (four numbers). Die B's even numbers: 2,4,4,6 (four numbers). So each odd number on Die A (4 numbers) paired with each even number on Die B (4 numbers) gives 4*4=16 pairs where sum is odd + even = odd.Then Die A's even numbers: 2,6 (two numbers). Die B's odd numbers:1,5 (two numbers). Each of these two on A paired with two on B gives 2*2=4 pairs where even + odd = odd.So total odd sums: 16 + 4 = 20. Total possible pairs: 6*6=36. Probability 20/36=5/9. So that's the same as before.Therefore, answer is D.I don't see any mistakes here, so I think 5/9 is the correct answer.Final Answerboxed{D}</think>To determine the probability that the sum of two modified six-sided dice is odd, we first analyze the composition of each die:- Die A (with 4 replaced by 3): The faces are 1, 2, 3, 3, 5, 6. This die has 4 odd numbers (1, 3, 3, 5) and 2 even numbers (2, 6). The probability of rolling an odd number is (frac{4}{6} = frac{2}{3}), and the probability of rolling an even number is (frac{2}{6} = frac{1}{3}).- Die B (with 3 replaced by 4): The faces are 1, 2, 4, 4, 5, 6. This die has 2 odd numbers (1, 5) and 4 even numbers (2, 4, 4, 6). The probability of rolling an odd number is (frac{2}{6} = frac{1}{3}), and the probability of rolling an even number is (frac{4}{6} = frac{2}{3}).The sum of the two dice is odd if one die shows an odd number and the other shows an even number. We calculate this probability as follows:- Probability that Die A is odd and Die B is even: (frac{2}{3} times frac{2}{3} = frac{4}{9})- Probability that Die A is even and Die B is odd: (frac{1}{3} times frac{1}{3} = frac{1}{9})Adding these probabilities gives the total probability of an odd sum:[frac{4}{9} + frac{1}{9} = frac{5}{9}]Thus, the probability that the sum is an odd number is (boxed{D}).,<think>Okay, let me try to figure out this problem. So we have two six-sided dice. Each face is equally likely, so each number has a 1/6 chance. But one die has the 4 replaced by 3, and the other die has the 3 replaced by 4. We need to find the probability that the sum is an odd number when these two dice are rolled. The options are given from A to E, so I need to calculate it step by step.First, let me understand the dice. Normally, a die has numbers 1 through 6. The first die has 4 replaced by 3. So instead of 1, 2, 3, 4, 5, 6, it has 1, 2, 3, 3, 5, 6. Let's confirm: original numbers 1,2,3,4,5,6; replacing 4 with 3 gives 1,2,3,3,5,6. So the first die has two 3s and no 4. Let me list the numbers on Die 1: 1, 2, 3, 3, 5, 6.The second die has the 3 replaced by 4. So original numbers 1,2,3,4,5,6; replacing 3 with 4 gives 1,2,4,4,5,6. Wait, so Die 2 has two 4s and no 3? Let's check: yes, instead of 3, it's 4, so numbers are 1,2,4,4,5,6.So Die 1: [1,2,3,3,5,6]Die 2: [1,2,4,4,5,6]Now, when we roll these two dice, the possible outcomes are all pairs (a, b) where a is from Die 1 and b is from Die 2. Since each die has 6 faces, there are 6*6=36 possible outcomes, each with probability 1/36.We need to find the number of outcomes where the sum a + b is odd, then divide by 36 to get the probability.So, the key idea here is that an odd sum occurs when one die is even and the other is odd. Because even + odd = odd. So if Die 1 is even and Die 2 is odd, or Die 1 is odd and Die 2 is even, then the sum is odd. If both are even or both are odd, the sum is even.Therefore, the probability that the sum is odd is equal to (number of pairs where one is even and the other is odd)/36.Therefore, we can compute this by calculating how many even and odd numbers are on each die, then using those counts to find the number of favorable pairs.So first, let's find the number of even and odd numbers on each die.Starting with Die 1: [1,2,3,3,5,6]Numbers on Die 1:1: odd2: even3: odd3: odd5: odd6: evenSo Die 1 has 2 even numbers (2 and 6) and 4 odd numbers (1, 3, 3, 5).Wait, hold on: 1,3,3,5 are four odd numbers, and 2,6 are two even numbers. So Die 1: 4 odd, 2 even.Now Die 2: [1,2,4,4,5,6]Numbers on Die 2:1: odd2: even4: even4: even5: odd6: evenSo Die 2: 2 odd numbers (1 and 5) and 4 even numbers (2,4,4,6).So Die 2: 2 odd, 4 even.Therefore, the counts are:Die1: 4 odd, 2 evenDie2: 2 odd, 4 evenSo the number of favorable outcomes where the sum is odd is equal to (number of odd on Die1 * number of even on Die2) + (number of even on Die1 * number of odd on Die2)Calculating that:Odd on Die1 * Even on Die2: 4 * 4 = 16Even on Die1 * Odd on Die2: 2 * 2 = 4Therefore, total favorable outcomes: 16 + 4 = 20Hence, probability is 20/36 = 5/9, which is option D.Wait, but let me check if that's correct. Wait, Die1 has 4 odd numbers, Die2 has 4 even numbers. So 4*4 = 16. Die1 has 2 even numbers, Die2 has 2 odd numbers. 2*2=4. 16+4=20. 20/36=5/9. So answer is D. Hmm, but let me cross-verify.Alternatively, maybe listing all possible pairs is possible, but that would be tedious. However, perhaps it's better to check if the parity (even or odd) is correct for each die.Wait, let's check the individual numbers on each die again:Die1: 1 (odd), 2 (even), 3 (odd), 3 (odd), 5 (odd), 6 (even). So as per earlier: 4 odd, 2 even.Die2: 1 (odd), 2 (even), 4 (even), 4 (even), 5 (odd), 6 (even). So 2 odd, 4 even. Correct.Therefore, when you compute the possible pairs where one is odd and the other even, you get (4*4)+(2*2)=16+4=20.Therefore, 20/36 simplifies to 5/9, which is D.But wait, let me check again. Maybe there's a mistake in the calculation? Because 4*4 is 16, and 2*2 is 4. 16+4=20. 20 divided by 36 is indeed 5/9, which is option D. So that seems right.But just to be thorough, let me think about the possible numbers each die can roll.Alternatively, maybe we can list all possible odd and even numbers for each die and check.Die1 numbers:1 (odd)2 (even)3 (odd)3 (odd)5 (odd)6 (even)So when rolled, probability of Die1 being odd is 4/6 = 2/3, and even is 2/6 = 1/3.Die2 numbers:1 (odd)2 (even)4 (even)4 (even)5 (odd)6 (even)So Die2: probability of odd is 2/6 = 1/3, even is 4/6 = 2/3.So the probability that Die1 is odd and Die2 is even is (2/3)*(2/3) = 4/9.Probability that Die1 is even and Die2 is odd is (1/3)*(1/3) = 1/9.Total probability: 4/9 + 1/9 = 5/9. Yep, same result. So 5/9, which is D. Therefore, the answer is D.But wait, hold on. Wait, the first calculation: Die1 odd (4/6) * Die2 even (4/6) = (4/6)(4/6)=16/36. Die1 even (2/6) * Die2 odd (2/6)= (2/6)(2/6)=4/36. So total 20/36=5/9. So same as before. So that's correct.Alternatively, perhaps there's an error in assuming that the number of even and odd numbers is the same as the counts. Let me think. For Die1, replacing a 4 with a 3. Original die has three even numbers (2,4,6) and three odd (1,3,5). But replacing 4 with 3, so now Die1 has even numbers 2,6 (since 4 is replaced by 3) and odd numbers 1,3,3,5. So yes, 2 even, 4 odd. Correct.Die2, replacing 3 with 4. Original die had three even (2,4,6) and three odd (1,3,5). Replacing 3 with 4: so now Die2 has even numbers 2,4,4,6 and odd numbers 1,5. So 4 even, 2 odd. Correct.Therefore, the counts are accurate. Therefore, the calculation is correct.Therefore, the answer is D, 5/9. So the answer is D.Final Answerboxed{D}</think>We have two six-sided dice. The first die has the 4 replaced by 3, resulting in faces [1, 2, 3, 3, 5, 6]. The second die has the 3 replaced by 4, resulting in faces [1, 2, 4, 4, 5, 6]. We need to find the probability that the sum of the numbers rolled on these dice is odd.To determine this, we first identify the number of odd and even faces on each die:- Die 1: Faces are [1, 2, 3, 3, 5, 6]. This die has 4 odd numbers (1, 3, 3, 5) and 2 even numbers (2, 6).- Die 2: Faces are [1, 2, 4, 4, 5, 6]. This die has 2 odd numbers (1, 5) and 4 even numbers (2, 4, 4, 6).An odd sum occurs when one die shows an odd number and the other shows an even number. We calculate the number of favorable outcomes:- Odd on Die 1 and Even on Die 2: (4 times 4 = 16)- Even on Die 1 and Odd on Die 2: (2 times 2 = 4)Total favorable outcomes: (16 + 4 = 20)The probability is the number of favorable outcomes divided by the total number of possible outcomes (36):[frac{20}{36} = frac{5}{9}]Thus, the probability that the sum is an odd number is (boxed{D}).]

🔑:On the first die, the chance of an odd number is frac{4}{6} = frac{2}{3}, and the chance of an even number is frac{1}{3}.On the second die, the chance of an odd number is frac{1}{3}, and the chance of an even number is frac{2}{3}.To get an odd sum, we need exactly one even and one odd.The odds of the first die being even and the second die being odd is frac{1}{3}cdotfrac{1}{3} = frac{1}{9}The odds of the first die being odd and the second die being even is frac{2}{3}cdotfrac{2}{3} = frac{4}{9}.The overall probability is frac{1}{9} + frac{4}{9} = frac{5}{9}, and the answer is boxed{D}.