Решение:

1. a) $$\frac{5x-1}{x^2} + \frac{x-3}{4x} = \frac{4(5x-1)}{4x^2} + \frac{x(x-3)}{4x^2} = \frac{20x-4+x^2-3x}{4x^2} = \frac{x^2+17x-4}{4x^2}$$
2. б) $$\frac{1}{3a-b} - \frac{1}{3a+b} = \frac{(3a+b)-(3a-b)}{(3a-b)(3a+b)} = \frac{3a+b-3a+b}{(3a)^2-b^2} = \frac{2b}{9a^2-b^2}$$
3. в) $$\frac{7}{c+3} - \frac{7c-5}{c^2+3c} = \frac{7}{c+3} - \frac{7c-5}{c(c+3)} = \frac{7c}{c(c+3)} - \frac{7c-5}{c(c+3)} = \frac{7c-7c+5}{c(c+3)} = \frac{5}{c(c+3)}$$

Ответ:

1. a) $$\frac{x^2+17x-4}{4x^2}$$
2. б) $$\frac{2b}{9a^2-b^2}$$
3. в) $$\frac{5}{c(c+3)}$$