Решение:

1)

a)

$$\frac{x}{3} + \frac{x-2}{5} = \frac{5x + 3(x-2)}{15} = \frac{5x + 3x - 6}{15} = \frac{8x - 6}{15}$$

Ответ: $$\frac{8x - 6}{15}$$

б)

$$\frac{3y-2}{6} - \frac{y+1}{4} = \frac{2(3y-2) - 3(y+1)}{12} = \frac{6y - 4 - 3y - 3}{12} = \frac{3y - 7}{12}$$

Ответ: $$\frac{3y - 7}{12}$$

в)

$$-\frac{b-c}{7} + \frac{3b-c}{14} = \frac{-2(b-c) + (3b-c)}{14} = \frac{-2b + 2c + 3b - c}{14} = \frac{b + c}{14}$$

Ответ: $$\frac{b + c}{14}$$

г)

$$\frac{1}{a^2} + \frac{a-2}{a} = \frac{1 + a(a-2)}{a^2} = \frac{1 + a^2 - 2a}{a^2} = \frac{a^2 - 2a + 1}{a^2} = \frac{(a-1)^2}{a^2}$$

Ответ: $$\frac{(a-1)^2}{a^2}$$

д)

$$\frac{3x-5}{x} - \frac{3y-2}{y} = \frac{y(3x-5) - x(3y-2)}{xy} = \frac{3xy - 5y - 3xy + 2x}{xy} = \frac{2x - 5y}{xy}$$

Ответ: $$\frac{2x - 5y}{xy}$$

e)

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

Ответ: $$\frac{(b-a)(b+a)}{ab^2}$$

2)

a)

$$\frac{(x+y)^2}{6y} + \frac{(x-y)^2}{12y} - \frac{x^2-y^2}{4y} = \frac{2(x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) - 3(x^2 - y^2)}{12y} = \frac{2x^2 + 4xy + 2y^2 + x^2 - 2xy + y^2 - 3x^2 + 3y^2}{12y} = \frac{2xy + 6y^2}{12y} = \frac{2y(x + 3y)}{12y} = \frac{x + 3y}{6}$$

Ответ: $$\frac{x + 3y}{6}$$

б)

$$\frac{3a+1}{7a} - \frac{7a+b}{14ab} - \frac{b-1}{2b} = \frac{2b(3a+1) - (7a+b) - 7a(b-1)}{14ab} = \frac{6ab + 2b - 7a - b - 7ab + 7a}{14ab} = \frac{-ab + b}{14ab} = \frac{b(1 - a)}{14ab} = \frac{1 - a}{14a}$$

Ответ: $$\frac{1 - a}{14a}$$