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若x-1=(y+1)/2=9z-2)/3,则x方+y方+z方的最小值是
题目详情
若x-1=(y+1)/2=9z-2)/3,则x方+y方+z方的最小值是
▼优质解答
答案和解析
x-1=(y+1)/2
y+1=2x-2
y=2x-3
x-1=(9z-2)/3
9z-2=3x-3
z=(3x-1)/9=x/3-1/9
所以x^2+y^2+z^2
=x^2+(2x-3)^2+(x/3-1/9)^2
=x^2+4x^2-12x+9+x^2/9-2x/27+1/81
=(46/9)x^2-(326/27)x+730/81
=(46/9)(x-163/138)^2+779/414
所以最小值=779/414
y+1=2x-2
y=2x-3
x-1=(9z-2)/3
9z-2=3x-3
z=(3x-1)/9=x/3-1/9
所以x^2+y^2+z^2
=x^2+(2x-3)^2+(x/3-1/9)^2
=x^2+4x^2-12x+9+x^2/9-2x/27+1/81
=(46/9)x^2-(326/27)x+730/81
=(46/9)(x-163/138)^2+779/414
所以最小值=779/414
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