a) ĐKXĐ: \(x\ne-1;0;1.\)Ta có:
\(A=\left[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\)
\(=\left[\frac{2}{\left(x+1\right)^3}\cdot\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}\cdot\frac{x^2+1}{x^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\left[\frac{2}{x\left(x+1\right)^2}+\frac{x^2+1}{x^2\left(x+1\right)^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\left[\frac{2x}{x^2\left(x+1\right)^2}+\frac{x^2+1}{x^2\left(x+1\right)^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\cdot\frac{x^3}{x-1}\)
\(=\frac{\left(x+1\right)^2\cdot x}{\left(x+1\right)^2\left(x-1\right)}=\frac{x}{x-1}.\)
Vậy \(A=\frac{x}{x-1}\)với \(x\ne-1;0;1.\)
b) A < 1 \(\Leftrightarrow\frac{x}{x-1}< 1\Leftrightarrow\frac{x}{x-1}-1< 0\Leftrightarrow\frac{x}{x-1}-\frac{x-1}{x-1}< 0\)\(\Leftrightarrow\frac{1}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)(do 1 > 0)\(\Leftrightarrow x< 1.\)
Kết hợp ĐKXĐ, A < 1 khi \(x< 1\)và \(x\ne-1;0.\)
c) \(A\inℤ\Leftrightarrow\frac{x}{x-1}\inℤ.\)Mà \(x\inℤ\)\(\Rightarrow x⋮\left(x-1\right)\Rightarrow\left(x-1+1\right)⋮\left(x-1\right)\Rightarrow1⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(1\right)=\left\{1;-1\right\}.\)Ta lập bảng sau:
| \(x-1\) | 1 | -1 |
| \(x\) | 2 | 0 |
| Kết luận | x thoả mãn ĐKXĐ | x không thoả mãn ĐKXĐ |
Vậy để A nguyên thì x = 2.
