Question

Phân tích thành nhân tử

a) (x+2)(x+3)(x-7)(x-8)-144

b) (x2+5x)2+10x2+5x+24

Answer 1

Ta có :\(\left(a+2\right)\left(x+3\right)\left(x+7\right)\left(x+8\right)-144\)

\(=\left(a+2\right)\left(x+7\right)\left(x+3\right)\left(x+8\right)-144\)

\(=\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\)

\(=\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\)

Đặt \(x^2-5x-14=t\) ta có:

\(t\left(t-10\right)-144\)

\(=t^2-10t-144\)

\(=\left(t-18\right)\left(t+8\right)\)

hay \(\left(x^2-5x-32\right)\left(x^2-5x-6\right)\)

\(=\left(x^2-5x-32\right)\left(x+1\right)\left(x-6\right)\)

b)