\(C=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) \(\left(1\right)\)
Đặt \(x^2+8x+11=t\) , khi đó
\(\left(1\right)\Leftrightarrow\left(t-4\right)\left(t+4\right)+15\)
\(=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\\ =\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
\(C=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(t=x^2+8x+7\) thì C trở thành:
\(t\left(t+8\right)+15=t^2+8t+15\)
\(t^2+3t+5t+15=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+5\right)\left(t+3\right)=\left(x^2+8x+7+5\right)\left(x^2+8x+7+3\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
\(C=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(C=\left(x^2+8+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=k\)
\(\Rightarrow C=k\left(k+8\right)+15=k^2+8k+15\)
\(\Rightarrow C=k^2+3k+5k+15\)
\(\Rightarrow C=k\left(k+3\right)+5\left(k+3\right)\)
\(\Rightarrow C=\left(k+3\right)\left(k+5\right)\)
\(\Rightarrow C=\left(x^2+8x+7+6\right)\left(x^2+8x+7+3\right)\)
\(\Rightarrow C=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(\Rightarrow C=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
