Question

sử dụng phương pháp đặt ẩn phụ

1, (x+1)(x+3)(x+5)(x+7) +15

Answer 1

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+7x+x+7\right)\left(x^2+5x+3x+15\right)+15\)

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

Đặt \(a=x^2+8x+11\) ta có

\(=\left(a-4\right)\left(a+4\right)+15\)

\(=a^2-16+15\)

\(=a^2-1\)

\(=\left(a-1\right)\left(a+1\right)\)

\(=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)