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x^2+y^2_4x+y+4.25=0求y^-x+3xy的值
人气:318 ℃ 时间:2020-09-10 07:00:09
解答
x^2+y^2-4x+y+4.25=0
x^2-4x+4+y^2+y+0.25=0
(x-2)^2+(y+0.5)^2=0
(x-2)^2=0,(y+0.5)^2=0
x=2
y=-0.5
y^-x+3xy
=(-0.5)^-2+3*2*(-0.5)
=(-2)^2-3
=4-3
=1
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