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已知x,y为实数,且x⒉+1/2y⒉+4≤xy+2y,求x,y的值?
人气:263 ℃ 时间:2020-05-23 17:58:21
解答
x^2+1/2y⒉+4≤xy+2y
x^2+1/2y^2+4-xy-2y<=0
(x-y/2)^2+y^2/4-2y+4<=0
(x-y/2)^2+(y/2-2)^2<=0
x-y/2=0, y/2-2=0
y=4, x=2.
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