设x2y2-20xy+x2+y2+81=0,求x+y的值

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设x2y2-20xy+x2+y2+81=0,求x+y的值
设x2y2-20xy+x2+y2+81=0,求x+y的值

设x2y2-20xy+x2+y2+81=0,求x+y的值
x2y2-20xy+x2+y2+81=0x2y2-18xy+81+x2-2xy+y2=0(XY-9)的平方+(X-Y)=0XY-9=0 X-Y=0第一种 X=-3 Y=-3 所以x+y=-3-3=-6
第二种 X=3 Y=3 所以x+y=3+3=6
于是x+y=正负6

3.∵x2y2-20xy+x2+y2+81=0,
∴x2y2-18xy+81+x2-2xy+y2=0,
∴(xy-9)2+(x-y)2=0,
∴xy-9=0且x-y=0,
∴当x=3时,y=3;当x=-3时,y=-3.

x2y2-20xy+x2+y2+81=0 x^2y^2-18xy+81+x^2+y^2-2xy=0 (xy-9)^2+(x+y)^2=0 xy=9且x-y=0 x=3,y=3 所以x=y=6

x2y2-20xy+x2+y2+81=0 x2y2-18xy+81+x2-2xy+y2=0(xy-9)²+(x-y)²=0 xy-9=0 x-y=0 x=-3 y=-3或 x=3 y=3 所以x+y=6或x+y=-6

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