$\begin{cases}y^2-2xy=8x^2-6x+1(*)\\y^2=x^3+8x^2-x+1(**)\end{cases}$
Ta có: $y=4x-1⇔ y=1-2x$
Thay vào $(**)$, ta có:
$y=4x−1(**)$
$⇔16x^2−8x+1=x^3+8x^2−x+1⇒x^3−8x^2−x=0$
$⇒$ \(\left[ \begin{array}{l}x=0\\x=4±\sqrt[]{15}\end{array} \right.\)
$y=1-2x$
$⇒$ \(\left[ \begin{array}{l}x=0\\x=-3\\x=-1\end{array} \right.\)
$............$