\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Rightarrow3x^2-8xy+4y^2=0\)
\(\Rightarrow\left(3x-2y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu...
\(\left\{{}\begin{matrix}2x^2-3xy+y^2=3\\x^2+2xy-2y^2=6\end{matrix}\right.\)\(\left(1\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Leftrightarrow3x^2-8xy+4y^2=0\)
\(\Leftrightarrow3x\left(x-2y\right)-2y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(3x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=\dfrac{2y}{3}\end{matrix}\right.\)
Thay vào \(\left(1\right)\) ta được:
\(\Leftrightarrow\left[{}\begin{matrix}2.\left(2y\right)^2-3.2y.y+y^2=3\\2.\left(\dfrac{2y}{3}\right)^2-3.\dfrac{2y}{3}.y+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2=1\\y^2=-27\left(VLý\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
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