nếu là lớp 8 thì rất hoan nghênh

a) $\Delta'=\left(\sqrt{3}\right)^2-3\cdot\left(-3\right)=12>0$

phương trình có 2 nghiệm phân biệt:

$\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+\sqrt{12}}{3}=\sqrt{3}\\x=\dfrac{\sqrt{3}-\sqrt{12}}{3}=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.$

kết luận: $x=\sqrt{3}$, $x=-\dfrac{\sqrt{3}}{3}$

b) $\left\{{}\begin{matrix}x\left(x-1\right)+y=\left(x+1\right)\left(x-3\right)\\2x-3y=-1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}y=\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\\2x-3\left(\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\right)=-1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}y=\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\\2x-3\left(-x-3\right)=-1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}y=\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\\5x+9=-1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.$

kết luận: $\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.$