Question

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

giải hệ pt

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\)

Trừ vế cho vế:

\(\Rightarrow3x^2-xy-2y^2=0\)

\(\Leftrightarrow\left(x-y\right)\left(3x+2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=-\dfrac{3}{2}x\end{matrix}\right.\)

Thế vào pt đầu: \(\left[{}\begin{matrix}2x^2-x^2=1\\2x^2-\left(-\dfrac{3}{2}\right)x^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=1\\-\dfrac{1}{4}x^2=1\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=y=1\\x=y=-1\end{matrix}\right.\)