\(x^2+2y+2xy+10x+12y+26=0\)
\(\Leftrightarrow\left[\left(x^2+2xy+y^2\right)+\left(10x+10y\right)+25\right]+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left[\left(x+y\right)^2+10\left(x+y\right)+25\right]+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x+y+5\right)^2+\left(y+1\right)^2=0\)
Vì \(\left(x+y+5\right)^2+\left(y+1\right)^2\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y+5=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-1\end{cases}}}\)
Vậy \(x=-4;y=-1\)
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