Sửa đề :
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt : \(y=x^2+5x+5\)
\(\Leftrightarrow\left(y-1\right)\left(y+1\right)-24=0\)
\(\Leftrightarrow y^2-25=0\)
\(\Leftrightarrow\left(y-5\right)\left(y+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-5=0\\y+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
+) \(y=5\Leftrightarrow x^2+5x+5=5\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
+) \(y=-5\Leftrightarrow x^2+5x+5=-5\)
\(\Leftrightarrow x^2+5x+10=0\)
\(\Leftrightarrow x^2+5x+\dfrac{25}{4}+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{15}{4}=0\) (vô lý)
Vậy....