Решение:

1. а) $$rac{3-2a}{2a} - \frac{1-a^2}{a^2} = \frac{a(3-2a)-2(1-a^2)}{2a^2} = \frac{3a-2a^2-2+2a^2}{2a^2} = \frac{3a-2}{2a^2}$$
2. б) $$rac{1}{3x+y} - \frac{1}{3x-y} = \frac{3x-y-(3x+y)}{(3x+y)(3x-y)} = \frac{3x-y-3x-y}{9x^2-y^2} = \frac{-2y}{9x^2-y^2}$$
3. в) $$rac{3}{b-2} - \frac{4-3b}{b^2-2b} = \frac{3}{b-2} - \frac{4-3b}{b(b-2)} = \frac{3b-(4-3b)}{b(b-2)} = \frac{3b-4+3b}{b(b-2)} = \frac{6b-4}{b(b-2)} = \frac{2(3b-2)}{b(b-2)}$$

Ответ:

• а) $$rac{3a-2}{2a^2}$$
• б) $$rac{-2y}{9x^2-y^2}$$
• в) $$rac{2(3b-2)}{b(b-2)}$$