\(A=\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)
\(=\left[x\left(x+1\right)+5\left(x+1\right)\right].\left[x\left(x+3\right)+7\left(x+3\right)\right]+15\)
\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right].\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=a\)
Ta có:
\(A=\left(a-4\right)\left(a+4\right)+15\)
\(=a^2-1=\left(a-1\right)\left(a+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right)\left[x\left(x+2\right)+6\left(x+2\right)\right]=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
Chúc bạn học tốt.
