Giải phương trình (x2-1)(x2+4x+3)=192

Đặt $t=x^2+2x-1$ , phương trình thành:

$\begin{array}{l}\left(t+2\right)\left(t-2\right)=192\iff t^2-4=192\iff t^2=196\\ \iff \left[\begin{array}{l}t=14\\ t=-14\end{array}\right.\iff \left[\begin{array}{l}{x}^2+2x-1=14\\ x^2+2x-1=-14\end{array}\right.\iff \left[\begin{array}{l}\left[\begin{array}{l}x=3\\ x=-5\end{array}\right.\\ \left[\begin{array}{l}x=-1+\sqrt{14}\\ x=-1-\sqrt{14}\end{array}\right.\end{array}\right.\end{array}$