Phương trình đã cho tương đương:
\(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\\ \Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\\ \Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-6\left(x^2+x\right)-24=0\\ \)
\(\Leftrightarrow\left(x^2+x\right)\left[\left(x^2+x\right)+4\right]-6\left[\left(x^2+x\right)+4\right]=24\\ \Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-6\right)=0\\ \Leftrightarrow\left(x^2+x+4\right)\left(x-2\right)\left(x+3\right)=0\)
Vì \(x^2+x+4>0\forall x\) nên x=2 và x=-3
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Rightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(t=x^2+x\) ta có:
\(t\left(t-2\right)-24=0\)\(\Rightarrow t^2-2t-24=0\)
\(\Rightarrow\left(t+4\right)\left(t-6\right)=0\)\(\Rightarrow\left[\begin{matrix}t=-4\\t=6\end{matrix}\right.\)
*)Xét \(t=-4\Rightarrow x^2+x=-4\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(loai\right)\)
*)Xét \(t=6\Rightarrow x^2+x=6\Rightarrow\left(x+3\right)\left(x-2\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
