lấy pt đầu trừ pt sau có:
$x^2-y^2=y-x$
$⇔(x-y)(x+y)+x-y=0$
$⇔(x-y)(x+y+1)=0$
$⇔\left[ \begin{array}{l}x=y\\x=-y-1\end{array} \right.$
th1: $x=y⇔y^2-y-6=0⇒\left[ \begin{array}{l}x=y=3\\x=y=-2\end{array} \right.$
th2: $x=-y-1$:
$(y+1)^2-y-6=0$
$⇔y^2+y-5=0$
$⇔y=\dfrac{-1±\sqrt[]{21}}{2}$
$⇒\left[ \begin{array}{l}x=\dfrac{-1+\sqrt[]{21}}{2}\\x=\dfrac{-1-\sqrt[]{21}}{2}\end{array} \right.$
