Решение:

а) $$\frac{x}{3} + \frac{x-2}{5} = \frac{5x + 3(x-2)}{15} = \frac{5x + 3x - 6}{15} = \frac{8x - 6}{15}$$ б) $$\frac{3y-2}{6} - \frac{y+1}{4} = \frac{2(3y-2) - 3(y+1)}{12} = \frac{6y - 4 - 3y - 3}{12} = \frac{3y - 7}{12}$$ в) $$\frac{b-c}{7} + \frac{3b-c}{14} = \frac{2(b-c) + (3b-c)}{14} = \frac{2b - 2c + 3b - c}{14} = \frac{5b - 3c}{14}$$ г) $$\frac{1}{a^2} + \frac{a-2}{a} = \frac{1 + a(a-2)}{a^2} = \frac{1 + a^2 - 2a}{a^2} = \frac{a^2 - 2a + 1}{a^2} = \frac{(a-1)^2}{a^2}$$ д) $$\frac{3x-5}{x} - \frac{3y-2}{y} = \frac{y(3x-5) - x(3y-2)}{xy} = \frac{3xy - 5y - 3xy + 2x}{xy} = \frac{2x - 5y}{xy}$$ e) $$\frac{b-a}{ab} - \frac{a-b}{b^2} = \frac{b(b-a) - a(a-b)}{ab^2} = \frac{b^2 - ab - a^2 + ab}{ab^2} = \frac{b^2 - a^2}{ab^2}$$

Ответ:

• а) $$\frac{8x - 6}{15}$$
• б) $$\frac{3y - 7}{12}$$
• в) $$\frac{5b - 3c}{14}$$
• г) $$\frac{(a-1)^2}{a^2}$$
• д) $$\frac{2x - 5y}{xy}$$
• е) $$\frac{b^2 - a^2}{ab^2}$$