\(A=\left(x^2+4x+8\right)^2-3x.\left(x^2+4x+8\right)+2x^2\)
Đặt \(t=x^2+4x+8\)
=> \(A=t^2-3tx+2x^2\)
\(=\left(t^2-tx\right)+\left(-2tx+2x^2\right)\)
\(=t\left(t-x\right)-2x\left(t-x\right)\)
\(=\left(t-x\right)\left(t-2x\right)\)
\(=\left(x^2+3x+8\right)\left(x^2+2x+8\right)\)
Không chắc lém :)))
Đặt (x2+4x+8)=t
Ta có :
t2-3xt+2x2
=t2-2xt-xt+2x2
=(t2-2xt)-(xt-2x2)
=t(t-2x)-x(t-2x)
=(t-2x)(t-x)
Vậy (x2+4x+8)2−3x.(x2+4x+8)+2x2
=(x2+4x+8-2x)(x2+4x+8-x)
=(x2+2x+8)(x2+3x+8)
