a, \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)\)
\(=\left[\left(x+2\right)\left(x-7\right)\right].\left[\left(x+3\right)\left(x-8\right)\right]\)
\(=\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\)(1)
Đặt \(x^2-5x-14=t\) thì \(x^2-5x-24=t-10\)
Thay vào (1), ta có:
\(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)\)
\(=t\left(t-10\right)-144\)
\(=t^2-10t-144\)
\(=t^2-18t+8t-144\)
\(=t\left(t-18\right)+8\left(t-18\right)\)
\(=\left(t+8\right)\left(t-18\right)\)
\(=\left(x^2-5x-14+8\right)\left(x^2-5x-14-18\right)\)
\(=\left(x^2-5x-6\right)\left(x^2-5x-32\right)\)
\(=\left(x+1\right)\left(x-6\right)\left(x^2-5x-32\right)\)