04.24. a) + (-1) a b b a(a-b b+a b² б) + 2a a(a-b);

a) $$\frac{b}{a} + \frac{b}{a(a-1)} = \frac{b(a-1) + b}{a(a-1)} = \frac{ba - b + b}{a(a-1)} = \frac{ba}{a(a-1)} = \frac{b}{a-1}$$ б) $$\frac{b+a}{2a} + \frac{b^2}{a(a-b)} = \frac{(b+a)(a-b) + 2b^2}{2a(a-b)} = \frac{ba - b^2 + a^2 - ab + 2b^2}{2a(a-b)} = \frac{a^2 + b^2}{2a(a-b)}$$ Ответ: a) $$\frac{b}{a-1}$$; б) $$\frac{a^2 + b^2}{2a(a-b)}$$