\(x^2+2y^2+2xy-2x+2y+5=0\)
\(\Leftrightarrow\left(x^2+2xy-2x+y^2-2y+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)^2+\left(y+2\right)^2=0\left(1\right)\)
Ta thấy: \(\begin{cases}\left(x+y-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\)
\(\Rightarrow\left(x+y-1\right)^2+\left(y+2\right)^2\ge0\left(2\right)\)
Từ (1) và (2) suy ra \(\begin{cases}\left(x+y-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x+y-1=0\\y+2=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x+y=1\\y=-2\end{cases}\)\(\Leftrightarrow\begin{cases}x-2=1\\y=-2\end{cases}\)\(\Leftrightarrow\begin{cases}x=3\\y=-2\end{cases}\)
Vậy các số x,y thỏa mãn là x=3; y=-2
\(x^2+2y^2+2xy-2x+2y+5=0\)
\(\Leftrightarrow x^2-2x+2xy+1-2y+y^2+y^2+4y+4=0\)
\(\Leftrightarrow x^2-2x\left(1-y\right)+\left(1-y\right)^2+y^2+4y+4=0\)
\(\Leftrightarrow\left(x+y-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left(x+y-1\right)=0and\left(y+2\right)^2=0\)
vậy x=3;y=-2
\(x^2+2y^2+2xy-2x+2y+5\)=0
\(\Leftrightarrow\)\(\left(x+y\right)^2-2\left(x+y\right)+1+y^2+4y+4\)=0
\(\Leftrightarrow\)\(\left(x+y-1\right)^2+\left(y+2\right)^2\)=0
\(\Rightarrow\left\{\begin{matrix}x+y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
