ĐK : \(x\ne0;-1;2\)
a) \(A=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(A=1+\left(\frac{x+1}{x^3+1}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(A=1+\frac{x+1+x+1-2\left(x^2-x+1\right)}{x^3+1}\cdot\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(A=1+\frac{-2x^2\left(x-2\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)\cdot x^2\left(x-2\right)}\)
\(A=1+\frac{-2}{x+1}\)
\(A=\frac{x-1}{x+1}\)
b) Để \(A\in Z\)\(\Leftrightarrow x-1⋮x+1\)
\(\Leftrightarrow x+1-2⋮x+1\)
\(\Leftrightarrow-2⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;1;-3\right\}\)( thỏa )
Vậy....
