b. 3 x − 1 4 + 1 ≥ x + 2 3

c) $\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2(x-11)}{x^2-4}$ ĐKXĐ: $x\ne \pm 2$

=> (x – 2)(x – 2) – 3(x+2)=2(x-11) = 0

$$\begin{gathered} \begin{array}{l}\iff x^2-4x+4-3x-6-2x+22=0\\ \iff x^2-9x+20=0\\ \iff x^2-4x-5x+20=0\\ \iff x(x-4)-5(x-4)=0\\ \iff (x-4)(x-5)=0\end{array} \\ \iff \left[\begin{array}{l}x=4\left(tm\right)\\ x=5\left(tm\right)\end{array}\right. \end{gathered}$$

Vậy: tập nghiệm của phương trình là: S = {4;5}

b) $\frac{1}{x+2}+\frac{5}{2-x}=\frac{2x-3}{x^2-4}$ điều kiện x $\ne \pm 2$

$\begin{array}{l}\iff \frac{1}{x+2}-\frac{5}{x-2}=\frac{2x-3}{x^2-4}\\ \iff x-2-5(x+2)=2x-3\\ \iff x-2-5x-10=2x-3\\ \iff -6x=9\iff x=\frac{-2}{3}\left(tmdk\right)\end{array}$

a) (2x - 1)(x + 2) = (3x - 2)(2x - 1)

<=> (2x-1)(x+2-3x+2) = 0

<=> (2x-1)(4-2x) = 0

<=> $x=\frac{1}{2}$ hoặc x = 2.

Vậy S = $\left\{\frac{1}{2};2\right\}$