\(\left(x-2\right)^2+2x-y=0\)\(\Leftrightarrow\left(x-1\right)^2+3-y=0\)
\(\Leftrightarrow\left(x-1\right)^2=y-3\Leftrightarrow\left[{}\begin{matrix}y-3=x-1\Leftrightarrow x=y-2\\y-3=1-x\Leftrightarrow x=4-y\end{matrix}\right.\)
chỉ giải ra được như thế thôi =))
\((x-2)^2+(2x-y)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3-y=0\)
\(\Leftrightarrow\left(x-1\right)^2=y-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=y-3\\x-1=3-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-2\\x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)