\(\left(x+4\right)\left(x+5\right)\left(x+7\right)\left(x+8\right)=4\)
\(\Leftrightarrow\left(x^2+12x+32\right)\left(x^2+12x+35\right)=4\)
Đặt \(x^2+12x+32=t\)
\(\Leftrightarrow t\left(t+3\right)=4\)
\(\Leftrightarrow\left(t-1\right)\left(t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-1=0\\t+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-4\end{matrix}\right.\)
Với \(t=1\)
\(\Leftrightarrow x^2+12x+32=1\)
\(\Leftrightarrow x^2+12x+31=0\)
\(\Delta'=6^2-31=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=-6+\sqrt{5}\\x=-6-\sqrt{5}\end{matrix}\right.\)
Với \(t=-4\)
\(\Leftrightarrow x^2+12x+32=-4\)
\(\Leftrightarrow\left(x+6\right)^2=0\)
\(\Leftrightarrow x=-6\)
Vậy \(S=\left\{-6;-6+\sqrt{5};-6-\sqrt{5}\right\}\)
