a: \(A=\frac{1}{x^2+x+1}+\frac{x^2+2}{x^3-1}\)
\(=\frac{1}{x^2+x+1}+\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x-1+x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{1}{x-1}\)
Thay x=11 vào A, ta được:
\(A=\frac{1}{11-1}=\frac{1}{10}\)
b: \(B=\frac{x+1}{x^2-x}+\frac{x+2}{1-x^2}\)
\(=\frac{x+1}{x\left(x-1\right)}-\frac{x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2-x\left(x+2\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1-x^2-2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{x\left(x-1\right)\left(x+1\right)}\)
Thay x=-1/3 vào B, ta được:
\(B=\frac{1}{-\frac13\left(-\frac13-1\right)\left(-\frac13+1\right)}=\frac{1}{-\frac13\cdot\frac{-4}{3}\cdot\frac23}=1:\frac{8}{27}=\frac{27}{8}\)
