若x^2-4x+y^2+6y+(z-3)开平方+13=0 求(xy)^2的值
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若x^2-4x+y^2+6y+(z-3)开平方+13=0 求(xy)^2的值
若x^2-4x+y^2+6y+(z-3)开平方+13=0 求(xy)^2的值
若x^2-4x+y^2+6y+(z-3)开平方+13=0 求(xy)^2的值
若x^2-4x+y^2+6y+(z-3)开平方+13=0
(x^2-4x+4)+(y^2+6y+9)+√(z-3)=0
(x-2)^2+(y+3)^2+√(z-3)=0
即
{x-2=0
{y+3=0
{z—3=0
解得
{x=2
{y=-3
{z=3
∴(xy)^2=(2×-3)^2=36
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