Đặt \(a=x^2+x+1\)\(\Rightarrow\)\(a+1=x^2+x+2\)
Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-6=a.\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=\left(a^2-2a\right)+\left(3a-6\right)\)
\(=a.\left(a-2\right)+3\left(a-2\right)\)
\(=\left(a+3\right).\left(a-2\right)\)
\(=\left(x^2+x+1+3\right).\left(x^2+x+1-2\right)\)
\(=\left(x^2+x+4\right)\left(x^2+x-1\right)\)
Chúc bn hok tốt
( x2 + x + 1 )( x2 + x + 2 ) - 6 (*)
Đặt x2 + x + 1 = t
(*) = t( t + 1 ) - 6
= t2 + t - 6
= t2 - 2t + 3t - 6
= t( t - 2 ) + 3( t - 2 )
= ( t - 2 )( t + 3 )
= ( x2 + x + 1 - 2 )( x2 + x + 1 + 3 )
= ( x2 + x - 1 )( x2 + x + 4 )
