Решение:

1. a) $$\frac{b}{a^2} - \frac{1}{a} = \frac{b}{a^2} - \frac{a}{a^2} = \frac{b-a}{a^2}$$
2. б) $$\frac{1-x}{x^2} + \frac{1}{x} = \frac{1-x}{x^2} + \frac{x}{x^2} = \frac{1-x+x}{x^2} = \frac{1}{x^2}$$
3. в) $$\frac{1}{2a^2} + \frac{4-3a^2}{2a^2} = \frac{1+4-3a^2}{2a^2} = \frac{5-3a^2}{2a^2}$$
4. г) $$\frac{a+b}{a^2b} + \frac{a-b}{ab^2} = \frac{b(a+b)}{a^2b^2} + \frac{a(a-b)}{a^2b^2} = \frac{ab+b^2+a^2-ab}{a^2b^2} = \frac{a^2+b^2}{a^2b^2}$$
5. д) $$\frac{2a-3b}{a^2b} - \frac{4a-5b}{ab^2} = \frac{b(2a-3b)}{a^2b^2} - \frac{a(4a-5b)}{a^2b^2} = \frac{2ab-3b^2-4a^2+5ab}{a^2b^2} = \frac{7ab-3b^2-4a^2}{a^2b^2}$$
6. е) $$\frac{x+2y}{xy^2} - \frac{2x-y}{x^2y} = \frac{x(x+2y)}{x^2y^2} - \frac{y(2x-y)}{x^2y^2} = \frac{x^2+2xy-2xy+y^2}{x^2y^2} = \frac{x^2+y^2}{x^2y^2}$$