\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Rightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)-24=0\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24=0\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+4=t\Rightarrow x^2+5x+6=t+2\) ta được:
\(t\left(t+2\right)-24=0\Rightarrow t^2+2t-24=0\)
\(\Rightarrow t^2-4t+6t-24=0\Rightarrow\left(t-4\right)\left(t-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t-4=0\\t-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=4\\t=6\end{matrix}\right.\)
Vì \(t=x^2+5x+4\) nên
\(\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+6=6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\x\left(x+5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy=-
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