\(\left\{{}\begin{matrix}x^2-3x+5y-2=0\\5x+y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3x+5\left(2-5x\right)-2=0\\y=2-5x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3x+10-25x-2=0\\y=2-5x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-28x+8=0\left(1\right)\\y=2-5x\end{matrix}\right.\)
Xét pt (1) ta có:
\(\Delta=\left(-28\right)^2-4\cdot1\cdot8=752>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{28+\sqrt{752}}{2}=14+2\sqrt{47}\\x_2=\dfrac{28-\sqrt{752}}{2}=14-2\sqrt{47}\end{matrix}\right.\)
Với \(x_1=14+2\sqrt{47}\Rightarrow y_1=2-5\cdot\left(14+2\sqrt{47}\right)=-68-10\sqrt{47}\)
Với \(x_2=14-2\sqrt{47}\Rightarrow y_2=2-5\cdot\left(14-2\sqrt{47}\right)=-68+10\sqrt{47}\)
Vậy: ...
