Giải các bất phương trình sau:
a) $\frac{2x^2-5x-6}{-3x+4}>-2$                      b) $\frac{1}{x+1}+\frac{1}{x-2} \geq \frac{1}{x-1}+\frac{1}{x}$
Giải
a) Ta có: $\frac{2x^2-5x-6}{-3x+4}>-2   (1)$
$\Leftrightarrow \frac{2x^2-5x-6}{-3x+4}+2>0 \Leftrightarrow \frac{2x^2-5x-6-6x+8}{-3x+4}>0$
$\Leftrightarrow \frac{2x^2-11x+2}{-3x+4}>0$
   Ta có: $2x^2-11x+2=0 \Leftrightarrow x=\frac{11\pm \sqrt{105}}{4}$
             $ -3x+4=0 \Leftrightarrow x=\frac{4}{3}$
Đặt $Q(x)=\frac{2x^2-11x+2}{-3x+4}$
Bảng xét dấu:

Vậy tập nghiệm của bất phương trình $(1)$ là: 
     $S=(-\infty;\frac{11-\sqrt{105}}{4})\cup (\frac{4}{3};\frac{11+\sqrt{105}}{4})$
b) Bất phương trình $(2)\Leftrightarrow \frac{1}{x+1}-\frac{1}{x-1} \geq \frac{1}{x}-\frac{1}{x-2}$
         $\Leftrightarrow \frac{-2}{x^2-1} \geq \frac{-2}{x^2-2x} \Leftrightarrow \frac{-2}{x^2-1}+\frac{2}{x^2-2x} \geq 0$
         $\Leftrightarrow \frac{-2(x^2-2x)+2(x^2-1)}{(x^2-1)(x^2-2x)} \geq 0 \Leftrightarrow \frac{4x-2}{(x^2-1)(x^2-2x)} \geq 0$
Lập bảng xét dấu của vế trái:

   Căn cứ vào bảng xét dấu, bất phương trình có tập nghiệm:
       $S=(-1;0)\cup [\frac{1}{2};1)\cup (2;+\infty)$
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