若x^2+4x+y^2-8y+20=0,则y^x的值是?
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若x^2+4x+y^2-8y+20=0,则y^x的值是?
若x^2+4x+y^2-8y+20=0,则y^x的值是?
若x^2+4x+y^2-8y+20=0,则y^x的值是?
x^2+4x+4+y^2-8y+16=0
(x+2)^2+(y-4)^2=0
x=-2 y=4
y^x=1/16
配方:(x+2)^2+(y-4)^2=0
所以x=-2 y=4
y^x=1/16
x^2+4x+y^2-8y+20=0,
即(x+2)~2+(y-4)~2=0即它们分别为0
所以x=-2 y=4
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