\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\left(1\right)\)
Đặt \(y=x^2+x+1\)
\(\left(1\right)=y\left(y+1\right)-12\)
\(=y^2+y-12\)
\(=y^2-3y+4y-12\)
\(=y\left(y-3\right)+4\left(y-3\right)\)
\(=\left(y-3\right)\left(y+4\right)\)
\(\Rightarrow\left(1\right)=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
Đặt \(t=x^2+x+1\) thì \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1-4\right)=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
