\(2x^2+y^2+2xy-8x-6y+10=0\)
\(\Rightarrow2.\left(2x^2+y^2+2xy-8x-6y+10\right)=0\)
\(\Rightarrow4x^2+2y^2+4xy-16x-12y+20=0\)
\(\Rightarrow\left(4x^2+y^2+16+4xy-8y-16x\right)+\left(y^2-4y+4\right)=0\)
\(\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2=0\left(1\right)\)
Ta có: \(\hept{\begin{cases}\left(2x+y-4\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2\ge0\forall x;y\left(2\right)}\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+y=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}2x+2=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
Chúc bạn học tốt.
