Question

tìm x, y:

\(4x^2\) +\(2y^2\) +\(2y\) -\(4xy\) +1=0

Answer 1

\(\Leftrightarrow4x^2-4xy+y^2+y^2+2y+1=0\)

\(\Leftrightarrow\left(2x-y\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=-1\end{matrix}\right.\)

Answer 2

\(4x^2+2y^2+2y-4xy+1=0\\ \Rightarrow\left(4x^2+y^2-4xy\right)+\left(y^2+2y+1\right)=0\\ \Rightarrow\left(2x-y\right)^2+\left(y+1\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}2x-y=0\\y+1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x=y\\y=-1\end{matrix}\right.\\ \Rightarrow x=\frac{-1}{2};y=-1\)