Question

(x+1)(x+3)(x+4)(x+6)-7

dùng phương pháp đặt ẩn phụ

 

Answer 1

\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)-7\)

\(=\left(x+1\right)\left(x+6\right)\left(x+3\right)\left(x+4\right)-7\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+12\right)-7\)

Đặt \(x^2+7x+9=t\)

\(=\left(t-3\right)\left(t+3\right)-7\)

\(=t^2-9-7=t^2-16=\left(t-4\right)\left(t+4\right)\)

\(=\left(x^2+7x+9-4\right)\left(x^2+7x+9+4\right)\)

\(=\left(x^2+7x+5\right)\left(x^2+7x+13\right)\)