\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\) , ta có :
\(\left(t-4\right)\left(t+4\right)+15\)
\(=t^2-16+15=t^2-1\)
\(=\left(t-1\right)\left(t+1\right)\)
\(=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
Ta có:A=(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x2+8x+7)(x2+8x+15)+15
Đăt:t=x2+8x+7
Khi đó:A=t(t+8)+15
=>A=t2+8t+15=>A=(t+3)(t+5)
hay A=(x2+8x+10)(x2+8x+12)
