\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left(a+1\right)\left(a+7\right).\left(a+3\right)\left(a+5\right)+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
Đặt : \(a^2+8a+11=b\) Khi đó A trở thành :
\(A=\left(b+4\right)\left(b-4\right)+15\)
\(\Leftrightarrow A=b^2-16+15\)
\(\Leftrightarrow A=b^2-1\)
\(\Leftrightarrow A=\left(b-1\right)\left(b+1\right)\)
Thay \(b=a^2+8a+11\) vào ta dc :
\(A=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)
\(\Leftrightarrow A=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)
\(\Leftrightarrow A=\left(a^2+8a+10\right)\left(a+2\right)\left(a+6\right)\)
Vậy....
