x^2+2y^2+4x-12y+22=0求x,y
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x^2+2y^2+4x-12y+22=0求x,y
2x^2
2x^2
![x^2+2y^2+4x-12y+22=0求x,y](/uploads/image/z/19384948-28-8.jpg?t=x%5E2%2B2y%5E2%2B4x-12y%2B22%3D0%E6%B1%82x%2Cy)
(x²+4x+4)+2(y²-6y+9)=0
(x+2)²+2(y-3)²=0
因为:(x+2)²≥0、(y-3)²≥0,则:
x+2=0且y-3=0
得:
x=-2、y=3
(x+2)²+2(y-3)²=0
因为:(x+2)²≥0、(y-3)²≥0,则:
x+2=0且y-3=0
得:
x=-2、y=3
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