a: \(B=\dfrac{x^2-1-2x+3x+1}{x\left(x-1\right)}=\dfrac{x^2+x}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
a) B = \(\dfrac{x+1}{x}-\dfrac{2}{x-1}+\dfrac{3x+1}{x\left(x-1\right)}\) (ĐK: \(x\ne0;1\))
= \(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}-\dfrac{2x}{x\left(x-1\right)}+\dfrac{3x+1}{x\left(x-1\right)}\)
= \(\dfrac{x^2-1-2x+3x+1}{x\left(x-1\right)}=\dfrac{x^2+x}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
b) \(\left|x\right|=1< =>\left[{}\begin{matrix}x=1\left(L\right)\\x=-1\left(C\right)\end{matrix}\right.\)
Thay x = -1 vào B, ta có:
\(\dfrac{-1+1}{-1-1}=0\)
c) B nguyên <=> \(\dfrac{x+1}{x-1}\) nguyên <=> \(1+\dfrac{2}{x-1}\) nguyên
<=> 2\(⋮x-1\)
<=> x-1 \(\in\left\{-2;-1;1;2\right\}\)
| x-1 | -2 | -1 | 1 | 2 |
| x | -1 | 0 | 2 | 3 |
| C | L | C | C |
KL: x \(\in\left\{-1;2;3\right\}\)