\(\left(x^2+6x+5\right)\left(x+4\right)\left(x+2\right)=40\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt \(x^2+6x+5=t\) ,ta có:
\(t\left(t+3\right)=40\)
\(\Leftrightarrow t^2+3t-40=0\)
\(\Leftrightarrow t^2+8t-5t-40=0\)
\(\Leftrightarrow\left(t+8\right)\left(t-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-8\\t=5\end{matrix}\right.\)
Với t = -8
\(x^2+6x+5=-8\)
\(\Leftrightarrow x^2+6x+13=0\) ( vô lý vì \(x^2+6x+13>0\forall x\) )
Với t = 5
\(x^2+6x+5=5\)
\(\Leftrightarrow x^2+6x=0\)
\(\Leftrightarrow x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy ............................
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