Question

Tìm x,y

x^2+2y^2+4x-2xy+4y+20=0

Answer 1

\(x^2+2y^2+4x-2xy+4y+20=0\)

\(x^2+y^2+y^2+4x-2xy+4y+20=0\)

\(\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+\left(4x+16\right)=0\)

\(\left(x-y\right)^2+\left(y+2\right)^2+4\left(x+4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y+2\right)^2=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\\x=-4\end{matrix}\right.\)

Mà \(-4-\left(-2\right)\ne0\)

=> Pt vô nghiệm