Giải các bất phương trình:
$a) |x-2|>|x+1|-3                  b) |\frac{3x-1}{1-5x} |<1$
a) Xét $x \ge 2 \Rightarrow \begin{cases}|x-2|=x-2 \\|x+1|=x+1 \end{cases}$. 
BPT $\Leftrightarrow x-2>x+1-3 \Leftrightarrow 0.x>0$ (BPT này vô nghiệm.)
    Xét $-1 < x < 2 \Rightarrow \begin{cases}|x-2|=2-x \\|x+1|=x+1\end{cases}$. 
BPT $\Leftrightarrow 2-x>x+1-3 \Leftrightarrow x<2$
   Xét $x \le -1 \Rightarrow \begin{cases}|x-2|=2-x \\|x+1|=-x-1 \end{cases}$. 
BPT $\Leftrightarrow 2-x>-x-1-3 \Leftrightarrow 6>0.x$ (BPT này nghiệm đúng với mọi $x \le -1$)
Vậy  $S=\left\{ { x \in \mathbb{R}| x <2} \right\}$.

b) Đưa bất phương trình về:   $-1<\frac{3x-1}{1-5x}<1  \Leftrightarrow    \begin{cases}\frac{3x-1}{1-5x}<1  \\\frac{3x-1}{1-5x}>-1   \end{cases} $
Kết quả:  Nghiệm của bất phương trình:  $x<0$  hoặc  $x>\frac{1}{4}$.

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