\(b,x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)\right].\left[\left(x-1\right)\left(x+2\right)\right]=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x=m\) , ta có :
\(m.\left(m-2\right)=24\)
\(\Leftrightarrow m^2-2m-24=0\)
\(\Leftrightarrow m^2-6m+4m-24=0\)
\(\Leftrightarrow m\left(m-6\right)+4\left(m-6\right)=0\)
\(\Leftrightarrow\left(m+4\right)\left(m-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m+4=0\\m-6=0\end{matrix}\right.\)
Nếu \(m+4=0\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=-\dfrac{15}{4}\)(vô lí, Loại)
Nếu \(m-6=0\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{2;-3\right\}\)
