$4x^2-12x+5=0$
\Rightarrow $4x^2-2x-10x+5=0$
\Rightarrow $(2x-1)(2x-5)=0$
\Rightarrow $\left[\begin{matrix} x=\dfrac{1}{2}\\ x=\dfrac{5}{2}\end{matrix}\right.$
$x(x-1)(x+1)(x+2)=24$
\Rightarrow $x(x+1)(x-1)(x+2)-24=0$
\Rightarrow $(x^2+x)(x^2+x-2)-24=0$
Đặt $x^2+x=t$
\Rightarrow $t(t-2)-24=0$
\Rightarrow $t^2-2t-24=0$
\Rightarrow $t^2-6t+4t-24=0$
\Rightarrow $(t-6)(t+4)=0$
\Rightarrow $(x^2+2-6)(x^2+2+4)=0$
\Rightarrow $(x-2)(x+3)(x^2+x+4)=0$
Do $x^2+x+4>0$
\Rightarrow $\left[\begin{matrix} x-2=0\\ x+3=0\end{matrix}\right.$
\Rightarrow $\left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.$