Đặt x2 + x +1 = k
Khi đó ta được: k(k + 1) - 12
= k2 + k - 12
= (k - 3)(k + 4)
Lại thay k = x2 + x +1 ta có đt:
(x2 + x + 1 - 3)(x2 + x + 1 + 4)
= (x2 + x - 2)(x2 + x +5).
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
Đặt \(x^2+x+1=t\) , ta có :
\(t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=\left(t^2+4t\right)-\left(3t+12\right)\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)