\(x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)+1=0\)
Đặt \(x^2+3x=a\) ta được:
\(a\left(a+2\right)+1=0\Leftrightarrow a^2+2a+1=0\)
\(\Rightarrow\left(a+1\right)^2=0\Rightarrow a=-1\Rightarrow x^2+3x=-1\)
\(\Rightarrow x^2+3x+1=0\Rightarrow x=\frac{-3\pm\sqrt{5}}{2}\)