\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
Đặt: \(a^2+8a+11=t\), khi đó pt trở thành:
\(\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)
\(=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\\ =\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)
\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
Đặt \(t=a^2+8a+7\) khi đó A thành:
\(t\left(t+8\right)+15=t^2+8t+15\)
\(=\left(t+3\right)\left(t+5\right)=\left(a^2+8a+7+3\right)\left(a^2+8a+7+5\right)\)
\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)
\(=\left(a^2+8a+10\right)\left(a+2\right)\left(a+6\right)\)
Ta có:
\(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(\left[\left(a+1\right)\left(a+7\right)\left(a+3\right)\left(a+5\right)\right]+15\)
\(\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
Đặt \(a^2+8a+7=t\)
\(\Rightarrow t\left(t+8\right)+15\)
\(=t^2+8t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(\Rightarrow\left[\left(a^2+8a+7\right)+3\right]\left[\left(a^2+8a+7\right)+5\right]\)
\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)
\(=\left(a^2+8a+10\right)\left(a^2+2a+6a+12\right)\)
\(=\left(a^2+8a+10\right)\left[a\left(a+2\right)+6\left(a+2\right)\right]\)
\(=\left(a^2+8a+10\right)\left(a+2\right)\left(a+6\right)\)
