\(x^2+10x+26+y^2+2y=0\)
\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Vậy \(x=-5\)và \(y=-1\)
\(x^2+10x+26+y^2+2y=0\)
\(\Leftrightarrow x^2+10x+25+y^2+2y+1=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Vậy..............
