Đề đúng phải là -24 chứ không phải +24
Ta có \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+11\)
\(\Rightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+24\left(1\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)+24\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)+24\)
Đặt \(k=x^2+7x+11\)
Thay vào ( 1 ) , ta có :
\(\left(k-1\right)\left(k+1\right)+24\)
\(\Leftrightarrow k^2-1^2+24\)
\(\Leftrightarrow k^2-25\)
\(\Leftrightarrow\left(k-5\right)\left(k+5\right)\)
\(\Leftrightarrow\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(\Leftrightarrow\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
