Pt \(\Leftrightarrow\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]-24=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x\right)-24=0\) (1)
Đặt \(a=x^2+x\) Khi đó pt(1) có dạng :
\(\left(a-2\right)a-24=0\)
\(\Leftrightarrow a^2-2a+1-25=0\)
\(\Leftrightarrow\left(a-1\right)^2-5^2=0\)
\(\Leftrightarrow\left(a-6\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=6\\a=-4\end{matrix}\right.\)
Với \(a=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) (t/m)
Với \(a=-4\Leftrightarrow x^2+x+4=0\) ( vô lí do VT > 0 )
Vậy pt đã cho có nghiệm \(S=\left\{2,-3\right\}\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)x\left(x+1\right)-24=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
