\(a,a^2-b^2+7a+7b\)
\(=\left(a^2-b^2\right)+\left(7a+7b\right)\)
\(=\left(a-b\right)\left(a+b\right)+7\left(a+b\right)\)
\(=\left(a+b\right)\left(a-b+7\right)\)
\(b,x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
a) Sửa đề: \(a^2-b^2+7a-7b\)
\(=\left(a-b\right)\left(a+b\right)+7\left(a-b\right)=\left(a-b\right)\left(a+b+7\right)\)
b) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)=\left(x-1\right)^2\left(x+1\right)\)
Vậy...