Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} 2x^2-xy=1(1)\\ (2x+y)^2=7(2)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left[{}\begin{matrix}2x+y=\sqrt{7}\\2x+y=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\sqrt{7}-2x\\y=-\sqrt{7}-2x\end{matrix}\right.\)
TH1: \(y=\sqrt{7}-2x\)
Thay vào pt đầu tiên:
\(\Rightarrow 2x^2-x(\sqrt{7}-2x)=1\)
\(\Leftrightarrow 4x^2-\sqrt{7}x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{23}+\sqrt{7}}{8}\rightarrow y=\dfrac{-\sqrt{23}+3\sqrt{7}}{4}\\x=\dfrac{-\sqrt{23}+\sqrt{7}}{8}\rightarrow y=\dfrac{\sqrt{23}+3\sqrt{7}}{4}\end{matrix}\right.\)
TH2: \(y=-\sqrt{7}-2x\)
Thay vào pt đầu tiên:
\(\Rightarrow 2x^2-x(-\sqrt{7}-2x)=1\)
\(\Leftrightarrow 4x^2+\sqrt{7}x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{23}-\sqrt{7}}{8}\rightarrow y=-\dfrac{\sqrt{23}+3\sqrt{7}}{4}\\x=\dfrac{-\sqrt{23}-7}{8}\rightarrow y=\dfrac{\sqrt{23}-3\sqrt{7}}{4}\end{matrix}\right.\)