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Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)
\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)
\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)
Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)
Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
\(A=\frac{2\left(x+1\right)}{x^3+1}=\frac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{2}{x^2-x+1}\)
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 \(x=\left\{0;1\right\}\)