\(4x^2+4xy+2y^2-4x-4y+2=0\)
\(\Rightarrow4x^2+4xy+y^2-4x-2y+1+y^2-2y+1=0\)
\(\Rightarrow\left(2x+1\right)^2-2\left(2x+1\right)+1+\left(y-1\right)^2=0\)
\(\Rightarrow\left(2x+1-1\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow4x^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}4x^2=0\\\left(y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}}\)
Đúng 0
Bình luận (0)