\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(Đặt\) \(t=x^2+5x+5\). PT thành
\(\left(t-1\right)\left(t+1\right)-24=0\Leftrightarrow t^2-25=0\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\Rightarrow\left[{}\begin{matrix}x^2+5x+5=5\\x^2+5x+5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\end{matrix}\right.\)
Vậy x=0 hoặc x=-5