Giải thích các bước giải:

$\begin{aligned}
& \text{a) } A = \dfrac{x}{x^2 - 5x + 6} - \dfrac{2}{2 - x} + \dfrac{x}{x-3} \\
& A = \dfrac{x}{x^2 - 2x - 3x + 6} + \dfrac{2}{x - 2} + \dfrac{x}{x-3} \\
& A = \dfrac{x}{x(x - 2) - 3(x - 2)} + \dfrac{2}{x - 2} + \dfrac{x}{x-3} \\
& A = \dfrac{x}{(x - 2)(x - 3)} + \dfrac{2(x - 3)}{(x - 2)(x - 3)} + \dfrac{x(x - 2)}{(x - 2)(x - 3)} \\
& A = \dfrac{x + 2x - 6 + x^2 - 2x}{(x - 2)(x - 3)} \\
& A = \dfrac{x^2 + x - 6}{(x - 2)(x - 3)} \\
& A = \dfrac{x^2 + 3x - 2x - 6}{(x - 2)(x - 3)} \\
& A = \dfrac{x(x + 3) - 2(x + 3)}{(x - 2)(x - 3)} \\
& A = \dfrac{(x - 2)(x + 3)}{(x - 2)(x - 3)} \\
& A = \dfrac{x + 3}{x - 3} \\
\\
& \text{b) } B = \dfrac{1 - 3x}{2x} + \dfrac{3x - 2}{2x - 1} + \dfrac{3x - 2}{2x - 4x^2} \\
& B = \dfrac{1 - 3x}{2x} + \dfrac{3x - 2}{2x - 1} - \dfrac{3x - 2}{4x^2 - 2x} \\
& B = \dfrac{1 - 3x}{2x} + \dfrac{3x - 2}{2x - 1} - \dfrac{3x - 2}{2x(2x - 1)} \\
& B = \dfrac{(1 - 3x)(2x - 1)}{2x(2x - 1)} + \dfrac{(3x - 2)2x}{2x(2x - 1)} - \dfrac{3x - 2}{2x(2x - 1)} \\
& B = \dfrac{2x - 1 - 6x^2 + 3x + 6x^2 - 4x - 3x + 2}{2x(2x - 1)} \\
& B = \dfrac{-2x + 1}{2x(2x - 1)} \\
& B = \dfrac{-(2x - 1)}{2x(2x - 1)} \\
& B = \dfrac{-1}{2x} \\
\\
& \text{c) } C = \dfrac{x}{2x - 2} + \dfrac{3x}{2x + 2} - \dfrac{2x^2}{x^2 - 1} \\
& C = \dfrac{x}{2(x - 1)} + \dfrac{3x}{2(x + 1)} - \dfrac{2x^2}{(x - 1)(x + 1)} \\
& C = \dfrac{x(x + 1)}{2(x - 1)(x + 1)} + \dfrac{3x(x - 1)}{2(x - 1)(x + 1)} - \dfrac{2x^2 \cdot 2}{2(x - 1)(x + 1)} \\
& C = \dfrac{x^2 + x + 3x^2 - 3x - 4x^2}{2(x - 1)(x + 1)} \\
& C = \dfrac{-2x}{2(x - 1)(x + 1)} \\
& C = \dfrac{-x}{(x - 1)(x + 1)} \\
& C = \dfrac{-x}{x^2 - 1} \\
& \text{Kết quả: a) } \dfrac{x + 3}{x - 3}; \text{ b) } \dfrac{-1}{2x}; \text{ c) } \dfrac{-x}{x^2 - 1}
\end{aligned}$