若x^2+2x+y^2-8y+17=0,求x^y的值

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若x^2+2x+y^2-8y+17=0,求x^y的值
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若x^2+2x+y^2-8y+17=0,求x^y的值
若x^2+2x+y^2-8y+17=0,求x^y的值

若x^2+2x+y^2-8y+17=0,求x^y的值
x^2+2x+y^2-8y+17=0
(x+1)^2+(y-4)^2=0
x+1=0,y-4=0
x=-1,y=4
x^y=(-1)^4=1

解由x^2+2x+y^2-8y+17=0,
得x^2+2x+1+y^2-8y+16=0,
即（x+1)^2+(y-4)^2=0
即x+1=0且y-4=0
即x=-1，y=4
即x^y
=（-1）^4
=1

解
x²+2x+y²-8y+17=0
(x²+2x+1)+(y²-8x+16)=0
(x+1)²+(y-4)²=0
由非负性知
x+1=0，y-4=0
∴x=-1，y=4
∴x^y
=(-1)^4
=1

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