若x-1=(y+1)/2=9z-2)/3,則x方+y方+z方的最小值是
若x-1=(y+1)/2=9z-2)/3,則x方+y方+z方的最小值是
其他人氣:767 ℃時(shí)間:2020-08-30 21:27:43
優(yōu)質(zhì)解答
x-1=(y+1)/2y+1=2x-2y=2x-3x-1=(9z-2)/39z-2=3x-3z=(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...
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