Question

\(\hept{\begin{cases}x^2-4xy+2y^2=1\\2x^2-xy+y^2=4\end{cases}}\)

Answer 1

\(\hept{\begin{cases}x^2-4xy+2y^2=1\\2x^2-xy+y^2=4\end{cases}}\)

\(\Leftrightarrow4.\left(x^2-4xy+2y^2\right)-2x^2+xy-y^2=0\)

\(\Leftrightarrow2x^2-15xy+7y^2=0\)

\(\Leftrightarrow2x^2-14xy-xy+7y^2=0\)

\(\Leftrightarrow2x.\left(x-7y\right)-y.\left(x-7y\right)=0\)

\(\Leftrightarrow\left(2x-y\right).\left(x-7y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=2x\\x=7y\end{cases}}\)

thay vào giải tiếp :]]]