2: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
Đặt \(x^2+x+1=a\)ta có
\(a\left(a+1\right)-12=a^2+a-12=a^2+4a-3a-12=a\left(a+4\right)-3\left(a+4\right)=\left(a+4\right)\left(a-3\right)\)
Thay \(a=x^2+x+1\)ta được
\(\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)Kl...
3. \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+7+8\right)+15\)
Đặt \(x^2+8x+7=a\) Ta có
\(a\left(a+8\right)+15=a^2+8a+15=a^2+5a+3a+15=a\left(a+5\right)+3\left(a+5\right)=\left(a+5\right)\left(a+3\right)\)
Thay \(a=x^2+8x+15\)ta được
\(\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)\)
