\(2x^2+y^2+4x-2y+3=0\)
\(\Leftrightarrow\left(2x^2+4x+2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow2.\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow2.\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\y-1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\y=1\end{array}\right.\)
Vậy \(x\in\left\{-1;1\right\}\)
Đúng 0
Bình luận (0)
