\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-12\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-12\)
Đăt \(a=x^2+5x+5\)
\(\Rightarrow x^2+5x+5=a-1\)(Trừ 1 cho 2 vế \(a=x^2+5x+5\))
\(\Rightarrow x^2+5x+6=a+1\)( Cộng 1 vào cả 2 vế \(a=x^2+5x+5\))
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-12=\left(a-1\right)\left(a+1\right)-12\)
\(=a^2-13\)
\(= \left(a-\sqrt{13}\right)\left(a+\sqrt{13}\right)\)
\(=\left(x^2+5x+5-\sqrt{13}\right)\left(x^2+5x+5+\sqrt{13}\right)\)