| 题号 | 答案 | 解析 |
| 1 | $\frac{5}{6}$ | 通分后：$\frac{4}{6} + \frac{1}{6} = \frac{5}{6}$ |
| 2 | $\frac{7}{8}$ | 通分后：$\frac{10}{8} - \frac{3}{8} = \frac{7}{8}$ |
| 3 | $\frac{9}{8}$ | 通分后：$\frac{4}{8} + \frac{6}{8} - \frac{1}{8} = \frac{9}{8}$ |
| 4 | $\frac{2}{9}$ | 先计算括号内：$\frac{6}{9} + \frac{1}{9} = \frac{7}{9}$，再减去：$\frac{7}{9} - \frac{7}{9} = \frac{2}{9}$ |
| 5 | $\frac{17}{20}$ | 通分后：$\frac{16}{20} + \frac{2}{20} - \frac{3}{20} = \frac{17}{20}$ |
| 6 | $\frac{5}{x}$ | 同分母相加：$\frac{3+2}{x} = \frac{5}{x}$ |
| 7 | $\frac{ad - bc}{bd}$ | 异分母通分后：$\frac{ad - bc}{bd}$ |
| 8 | $\frac{3x + 1}{x^2 - 1}$ | 通分后：$\frac{(x-1) + 2(x+1)}{(x+1)(x-1)} = \frac{3x + 1}{x^2 - 1}$ |
| 9 | $\frac{x^2 + x + 2}{(x-2)(x+2)}$ | 通分后：$\frac{x(x+2) - (x-2)}{(x-2)(x+2)} = \frac{x^2 + x + 2}{(x-2)(x+2)}$ |
| 10 | $\frac{2 + x - 2}{(x+2)(x-2)} = \frac{x}{(x+2)(x-2)}$ | 化简后：$\frac{2 + x - 2}{(x+2)(x-2)} = \frac{x}{(x+2)(x-2)}$ |