Question

Tìm x,y biết:

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

Answer 1

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

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

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

\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2\ge0\)

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

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

Vậy x = -1, y = 1