Ta có:\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt\(x^2+7x+10=t\)
\(=t\left(t+2\right)-24=t^2+2t-24=t^2+2t+1-25\)
\(=\left(t+1\right)^2-25=\left(t+6\right)\left(t-4\right)\)
Thay \(t=x^2+7x+10\) vào BT:
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)=\left(x^2+7x+16\right)\left(x+6\right)\left(x+1\right)\)