\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+10\) ta có:
\(=t\left(t+2\right)-24=t^2+2t-24\)
\(=t^2-4t+6t-24\)\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t-4\right)\left(t+6\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
(x+2)(x+3)(x+4)(x+5)-24
=(x^2+7x+10)(x^2+7x+12)-24
Đặt x^2+7x+10=a
a(a+2)-24
=a^2+2a-24
=(a-4)(a+6)
=(x^2+7x+6)(x^2+7x+16)
=(x+1)(x+6)(x^2+7x+16)
