Ta có \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\text{[}\left(x+2\right)\left(x+5\right)\text{]}.\text{[}\left(x+3\right)\left(x+4\right)\text{] -24}\)
\(=\left(x^2+7x+10\right)\times\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(x^2+7x+12=y+2\)
\(\Rightarrow y\left(y+2\right)-24=y^2+2y-24\)
\(=y^2-4y+6y-24\)
\(=y\left(y-4\right)+6\left(y-4\right)\)
\(=\left(y-4\right)\left(y+6\right)\left(1\right)\)
Thay\(y=x^2+7x+10\) vào (1) ta được
\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\text{[}x\left(x+1\right)+6\left(x+1\right)\text{]}\left(x^2+7x+6\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+6\right)\)
Vậy \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+6\right)\)
