Решение:
22.
- \( \frac{2a}{4a^2} = \frac{2a}{2a \cdot 2a} = \frac{1}{2a} \)
- \( \frac{5b}{10b^3} = \frac{5b}{5b \cdot 2b^2} = \frac{1}{2b^2} \)
23.
- \( \frac{4a^2b}{10a^3b^2} = \frac{2 \cdot 2a^2b}{2 \cdot 5a^2b \cdot ab} = \frac{2}{5ab} \)
- \( \frac{2x^2y^3}{10xy^4} = \frac{2x^2y^3}{5 \cdot 2xy^3 \cdot y} = \frac{x}{5y} \)
24.
- \( \frac{3x - 3y}{6x} = \frac{3(x-y)}{3 \cdot 2x} = \frac{x-y}{2x} \)
- \( \frac{10m - 5n}{5m} = \frac{5(2m-n)}{5m} = \frac{2m-n}{m} \)
25.
- \( \frac{am}{an+am} = \frac{am}{a(n+m)} = \frac{m}{n+m} \)
- \( \frac{mx-my}{m} = \frac{m(x-y)}{m} = x-y \)
Ответ:
22. а) \( \frac{1}{2a} \); б) \( \frac{1}{2b^2} \)
23. а) \( \frac{2}{5ab} \); б) \( \frac{x}{5y} \)
24. а) \( \frac{x-y}{2x} \); б) \( \frac{2m-n}{m} \)
25. а) \( \frac{m}{n+m} \); б) \( x-y \)