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已知x^2+y^2-4x+y+17\4=0,求y^-x+3xy的值.
人气:216 ℃ 时间:2020-04-03 15:25:51
解答
x^2+y^2-4x+y+17\4=0
x^2-4x+y^2+y+4+1/4=0
(x^2-4x+4)+(y^2+y+1/4)=0
(x-2)^2+(y+1/2)^2=0
∴(x-2)^2=0 (y+1/2)^2=0
∵平方是非负数
∴x-2=0 y+1/2=0
∴x=2 y=-1/2
∴y^-x+3xy
=-1/2^-2+3×2×-1/2
=4-3
=1
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