\(A=\frac{2\left(x+1\right)}{x^3+1}=\frac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{2}{x^2-x+1}\)
Để A nhận GT nguyên \(\Leftrightarrow x^2-x+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Mà \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\) nên
\(\orbr{\begin{cases}x^2-x+1=0\\x^2-x+1=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x\left(x-1\right)=0\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x-1\right)=0\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)x=0\\x-\frac{1}{2}=+-\sqrt{\frac{5}{4}}\left(l\right)\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x=\left\{0;1\right\}\)
