1/A= \(2\left(x^2+x\right)^2-9\left(x^2+x\right)+7\)
Đặt \(x^2+x=t\).Ta được:
A=\(2t^2-9t+7=\left(2t-7\right)\left(t-1\right)\)
\(=\left(2x^2+2x-7\right)\left(x^2+x-1\right)\)
2)B= \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+16\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16\)
Đặt \(x^2+8x=t\).Suy ra:
\(B=\left(t+7\right)\left(t+15\right)+16=t^2+22t+121\)
\(=\left(t+11\right)^2=\left(x^2+8x+11\right)^2\)
Đúng không ta?