已知x,y∈R且3x^2+2y^2=6x,求x+y的最大值与最小值答案3(x^2-2x+1)+2y^2=33(x-1)^2+2y^2=3(x-1)^2+2y^2/3=1令x-1=cosa,x=1+cosa则2y^2/3=1-cos²a=sin²a所以y=√(3/2)*sina所以x+y=1+cosa+√(3/2)*sina=√[(√3/2)^2+1^2]sin(a+z)+1=
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![已知x,y∈R且3x^2+2y^2=6x,求x+y的最大值与最小值答案3(x^2-2x+1)+2y^2=33(x-1)^2+2y^2=3(x-1)^2+2y^2/3=1令x-1=cosa,x=1+cosa则2y^2/3=1-cos²a=sin²a所以y=√(3/2)*sina所以x+y=1+cosa+√(3/2)*sina=√[(√3/2)^2+1^2]sin(a+z)+1=](/uploads/image/z/4001802-42-2.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%E2%88%88R%E4%B8%943x%5E2%2B2y%5E2%3D6x%2C%E6%B1%82x%2By%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%8E%E6%9C%80%E5%B0%8F%E5%80%BC%E7%AD%94%E6%A1%883%28x%5E2-2x%2B1%29%2B2y%5E2%3D33%28x-1%29%5E2%2B2y%5E2%3D3%28x-1%29%5E2%2B2y%5E2%2F3%3D1%E4%BB%A4x-1%3Dcosa%2Cx%3D1%2Bcosa%E5%88%992y%5E2%2F3%3D1-cos%26%23178%3Ba%3Dsin%26%23178%3Ba%E6%89%80%E4%BB%A5y%3D%E2%88%9A%283%2F2%29%2Asina%E6%89%80%E4%BB%A5x%2By%3D1%2Bcosa%2B%E2%88%9A%283%2F2%29%2Asina%3D%E2%88%9A%5B%28%E2%88%9A3%2F2%29%5E2%2B1%5E2%5Dsin%28a%2Bz%29%2B1%3D)
已知x,y∈R且3x^2+2y^2=6x,求x+y的最大值与最小值答案3(x^2-2x+1)+2y^2=33(x-1)^2+2y^2=3(x-1)^2+2y^2/3=1令x-1=cosa,x=1+cosa则2y^2/3=1-cos²a=sin²a所以y=√(3/2)*sina所以x+y=1+cosa+√(3/2)*sina=√[(√3/2)^2+1^2]sin(a+z)+1=
已知x,y∈R且3x^2+2y^2=6x,求x+y的最大值与最小值
答案3(x^2-2x+1)+2y^2=3
3(x-1)^2+2y^2=3
(x-1)^2+2y^2/3=1
令x-1=cosa,x=1+cosa
则2y^2/3=1-cos²a=sin²a
所以y=√(3/2)*sina
所以x+y=1+cosa+√(3/2)*sina
=√[(√3/2)^2+1^2]sin(a+z)+1
=√(5/2)sin(a+z)+1
所以最大值=(√10)/2+1,最小值=-(根号10)/2+1.不明白为什么要令x-1=cosa,x=1+cosa,2y^2/3=1-cos²a=sin²a
这样变了,整个题目就没变吗
已知x,y∈R且3x^2+2y^2=6x,求x+y的最大值与最小值答案3(x^2-2x+1)+2y^2=33(x-1)^2+2y^2=3(x-1)^2+2y^2/3=1令x-1=cosa,x=1+cosa则2y^2/3=1-cos²a=sin²a所以y=√(3/2)*sina所以x+y=1+cosa+√(3/2)*sina=√[(√3/2)^2+1^2]sin(a+z)+1=
这是一种方法
是参数方程的转化
只是未知量变化了,其实题目是没有变的.
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