Решение:

1) $$\frac{3}{2x} - \frac{2}{3x} = \frac{3 \cdot 3}{2x \cdot 3} - \frac{2 \cdot 2}{3x \cdot 2} = \frac{9}{6x} - \frac{4}{6x} = \frac{9-4}{6x} = \frac{5}{6x}$$

Ответ: $$\frac{5}{6x}$$

2) $$\frac{17y}{24c} - \frac{25y}{36c} = \frac{17y \cdot 3}{24c \cdot 3} - \frac{25y \cdot 2}{36c \cdot 2} = \frac{51y}{72c} - \frac{50y}{72c} = \frac{51y - 50y}{72c} = \frac{y}{72c}$$

Ответ: $$\frac{y}{72c}$$

3) $$\frac{b+2}{15b} - \frac{3c-5}{45c} = \frac{(b+2) \cdot 3c}{15b \cdot 3c} - \frac{(3c-5) \cdot b}{45c \cdot b} = \frac{3bc + 6c}{45bc} - \frac{3bc - 5b}{45bc} = \frac{3bc + 6c - (3bc - 5b)}{45bc} = \frac{3bc + 6c - 3bc + 5b}{45bc} = \frac{6c + 5b}{45bc}$$

Ответ: $$\frac{6c + 5b}{45bc}$$

4) $$\frac{8b+y}{40b} - \frac{6y+b}{30y} = \frac{(8b+y) \cdot 3y}{40b \cdot 3y} - \frac{(6y+b) \cdot 4b}{30y \cdot 4b} = \frac{24by + 3y^2}{120by} - \frac{24by + 4b^2}{120by} = \frac{24by + 3y^2 - (24by + 4b^2)}{120by} = \frac{24by + 3y^2 - 24by - 4b^2}{120by} = \frac{3y^2 - 4b^2}{120by}$$

Ответ: $$\frac{3y^2 - 4b^2}{120by}$$

5) $$\frac{1}{2a^7} + \frac{4-2a^3}{a^{10}} = \frac{1 \cdot a^3}{2a^7 \cdot a^3} + \frac{(4-2a^3) \cdot 2}{a^{10} \cdot 2} = \frac{a^3}{2a^{10}} + \frac{8 - 4a^3}{2a^{10}} = \frac{a^3 + 8 - 4a^3}{2a^{10}} = \frac{8 - 3a^3}{2a^{10}}$$

Ответ: $$\frac{8 - 3a^3}{2a^{10}}$$

6) $$\frac{a+b}{a^2} + \frac{a-b}{ab} = \frac{(a+b) \cdot b}{a^2 \cdot b} + \frac{(a-b) \cdot a}{ab \cdot a} = \frac{ab + b^2}{a^2b} + \frac{a^2 - ab}{a^2b} = \frac{ab + b^2 + a^2 - ab}{a^2b} = \frac{b^2 + a^2}{a^2b}$$

Ответ: $$\frac{b^2 + a^2}{a^2b}$$

7) $$\frac{7x+4}{8y} - \frac{3x-1}{6y} = \frac{(7x+4) \cdot 3}{8y \cdot 3} - \frac{(3x-1) \cdot 4}{6y \cdot 4} = \frac{21x + 12}{24y} - \frac{12x - 4}{24y} = \frac{21x + 12 - (12x - 4)}{24y} = \frac{21x + 12 - 12x + 4}{24y} = \frac{9x + 16}{24y}$$

Ответ: $$\frac{9x + 16}{24y}$$