Giải các bất phương trình:
a) $(1-7x)^2<(3+2x)^2$                                    b) $ \frac{x}{x^2+5x+6} \geq \frac{2}{x^2+3x+2}$
c) $\frac{x-2}{3x-1}<\frac{x+2}{2x-1}  $ 
d)  $\frac{x+7}{x^2+6x-7}+\frac{x-10}{x^2-13x+30} <0   $.
a) Đáp số:  $-\frac{2}{9}<x<\frac{4}{5}  $.
b) Ta có:    $x^2+5x+6=(x+2)(x+3)$
                   $x^2+3x+2=(x+1)(x+1)$.
Điều kiện xác định:  $x \neq -3; x \neq -2; x \neq -1$
Mẫu chung:  $(x+1)(x+2)(x+3)$  chuyển tất cả về vế trái, để vế phải bằng $0$,  quy đồng mẫu rút gọn, ta được bất phương trình
                  $\frac{x-3}{(x+1)(x+3)} \geq 0 $
Đáp số:  $x \ge 3;     -3<x<-1.$
c) Hướng dẫn:  sau khi thực hiện các phép  biến đổi, ta được bất phương trình:  $\frac{-x^2-10x+4}{(3x-1)(2x-1)}<0 $. 
     Kết quả:  $x <-5-\sqrt{29};  \frac{1}{3}<x<-5+\sqrt{29};  x>\frac{1}{2}  $.
d) Chú ý rằng :
$x^2+6x-7=(x+7)(x-1) \\ x^2-13x+30=(x-3)(x-10) $
BPT $\Leftrightarrow \frac{2(x-2)}{(x-1)(x-3)} < 0 \Leftrightarrow \left[ {\begin{matrix}x<1\\  2<x<3\end{matrix}} \right. $

Có thể tham khảo cách lập bảng xét dấu tại  Bài 107335.
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