\(A=\frac{1}{x^2-x}+\frac{1}{x^2+x+1}+\frac{2x}{1-x^3}\)
\(A=\frac{1}{x.\left(x-1\right)}+\frac{1}{x^2+x+1}+\frac{2x}{\left(1-x\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-x}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1+x^2-x-2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x^3-1\right)}\)
Với x=10
\(\Rightarrow A=\frac{1}{10.\left(10^3-1\right)}\)
\(A=\frac{1}{10.999}\)
\(A=\frac{1}{9990}\)
Vậy \(A=\frac{1}{9990}\)tại x=10
