Question

Tổng x+y,biết x,y thõa mãn x²+2y²-2xy+4y+4=0

Answer 1

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

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

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

Dễ thấy: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2\ge0\)

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