\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=3\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)=3\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=3\)
Đặt: \(x^2+5x=t\) , ta có:
\(\left(t+4\right)\left(t+6\right)=3\)
\(\Leftrightarrow t^2+10t+24=3\)
\(\Leftrightarrow t^2+10t+21=0\)
\(\Leftrightarrow\left(t+3\right)\left(t+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t+3=0\\t+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=-7\end{matrix}\right.\)
Suy ra, ta có:
\(\left[{}\begin{matrix}x^2+5x=-3\\x^2+5x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=0\\x^2+5x+7=0\end{matrix}\right.\)
Phần sau bạn tự làm nhá! Mình liệt rồi 
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