\(2x^2+10xy+14y^2+2x+2y+2=0\)
\(\Leftrightarrow\left(x^2+4y^2+1+2x+4xy+4y\right)+\left(x^2+6xy+9y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+2y+1\right)^2\ge0;\forall x,y\\\left(x+3y\right)^2\ge0;\forall x,y\\\left(y-1\right)^2\ge0;\forall x,y\end{cases}}\)
\(\Rightarrow\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2\ge0;\forall x,y\)
Do đó :\(\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+2y+1\right)^2=0\\\left(x+3y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=1\\x=-3\\y=1\end{cases}}\)
Vậy x=-3 và y=1
Kiến thức bổ sung
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow4x^2+20xy+28y^2+4x+4y+4=0\)
\(\Leftrightarrow\left(4x^2+4x+20xy+25y^2+10y+1\right)+\left(3y^2-6y+3\right)=0\)
\(\Leftrightarrow\left(2x+5y+1\right)^2+3\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+5y+1=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
