\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(t=x^2+x+1\) thì ta có:
\(t\left(t+1\right)-12=t^2+t-12\)
\(=t^2-3t+4t-12\)
\(=t\left(t-3\right)+4\left(t-3\right)\)\(=\left(t-3\right)\left(t+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
đặt \(t=x^2+x+1,5\)
khi đó:
\(BT=\left(t-0,5\right)\left(t+0,5\right)-12=t^2-0,25-12\\ =t^2-12,5=\left(t-\dfrac{5\sqrt{2}}{2}\right)\left(t+\dfrac{5\sqrt{2}}{2}\right)\\ =\left(x^2+x+1,5-\dfrac{5\sqrt{2}}{2}\right)\left(x^2+x+1,5+\dfrac{5\sqrt{2}}{2}\right)\)
