Question

sử dụng định lý bêzu

(x+1)(x+3)(x+5)(x+7)-9

Answer 1

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)-9\)

\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)-9\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9\)

Đặt \(x^2+8x+11=t\)

\(=\left(t-4\right)\left(t+4\right)-9\)

\(=t^2-16-9\)

\(=t^2-25=\left(t-5\right)\left(t+5\right)\)

Vậy \(=\left(x^2+8x+11-5\right)\left(x^2+8x+11+5\right)\)

\(=\left(x^2+8x+6\right)\left(x^2+8x+16\right)\)