\(2x^2+2y^2=5xy\Leftrightarrow2x^2-5xy+2y^2=0\Leftrightarrow2x^2-xy-4xy+2y^2=0\Leftrightarrow x\left(2x-y\right)-2y\left(2x-y\right)=0\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-2y=0\\2x-y=0\end{matrix}\right.\)
TH1:x-2y=0\(\Leftrightarrow x=2y\)
Đặt x=2k(k∈R)\(\Leftrightarrow\)y=k
Vậy \(\left\{{}\begin{matrix}x=2k\\y=k\end{matrix}\right.\)
TH2:2x-y=0\(\Leftrightarrow2x=y\)
Đặt y=2h(h∈R)\(\Leftrightarrow x=h\)
Vậy \(\left\{{}\begin{matrix}x=h\\y=2h\end{matrix}\right.\)
Vậy ta có các nghiệm tổng quát \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2k\\y=k\end{matrix}\right.\\\left\{{}\begin{matrix}x=h\\y=2h\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow2x^2-5xy+2y^2=0\)
\(\Leftrightarrow2x^2-4xy-xy+2y^2=0\)
\(\Leftrightarrow2x\left(x-2y\right)-y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\)
đến đây thì bí không biết làm nữa.
