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求(x-y)^2+(x-4)^2+(y+2)^2最小值
人气:418 ℃ 时间:2020-06-14 17:58:01
解答

(x-y)²+(x-4)²+(y+2)²
=(x-y)²/1+(4-x)²/1+(y+2)²/1
≥[(x-y)+(4-x)+(y+2)]²/(1+1+1)
=12.
故所求最小值为:12.
此时,x-y=4-x=y+2,
即x=2,y=0.
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