$a$) Đặt $t=x^2$
$⇒$ $t^2 - 3t - 4=0$
$⇔ t^2 + t - 4t - 4 = 0$
$⇔ t(t+1) - 4(t+1) = 0$
$⇔ (t-4)(t+1)=0$
$⇒$ \(\left[ \begin{array}{l}t=4\\t=-1\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x^2=4⇔ x = ± 2\\x^2=-1(KTM\end{array} \right.\)
Vậy $x$ $∈$ `{±2}`
$b$) $\left\{\begin{matrix}x+4y=0& \\3x-2y=7& \end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}3x+12y=0& \\3x-2y=7& \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix}3x+12y - (3y - 2y)=-7& \\3x-2y=7& \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix}14y=-7& \\3x-2y=7& \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix}y= \dfrac{-1}{2}& \\x=0-\dfrac{-4}{2}=2& \end{matrix}\right.$
