Решение:
- 1) \( \left(\frac{1}{x} + 2x\right) \cdot \frac{x^3}{4} = \left(\frac{1 + 2x^2}{x}\right) \cdot \frac{x^3}{4} = \frac{(1+2x^2)x^2}{4} = \frac{x^2+2x^4}{4} \)
- 2) \( \left(3y - \frac{1}{y}\right) \cdot \frac{y^4}{3} = \left(\frac{3y^2-1}{y}\right) \cdot \frac{y^4}{3} = \frac{(3y^2-1)y^3}{3} = \frac{3y^5 - y^3}{3} \)
- 3) \( \left(\frac{2a}{1+a} - 1\right) : \frac{a}{1+a} = \left(\frac{2a - (1+a)}{1+a}\right) : \frac{a}{1+a} = \left(\frac{a-1}{1+a}\right) \cdot \frac{1+a}{a} = \frac{a-1}{a} \)
- 4) \( \left(\frac{3}{b} + \frac{b}{b-1}\right) : \frac{b}{b-1} = \left(\frac{3(b-1) + b^2}{b(b-1)}\right) : \frac{b}{b-1} = \left(\frac{3b - 3 + b^2}{b(b-1)}\right) \cdot \frac{b-1}{b} = \frac{b^2+3b-3}{b^2} \)
Ответ: 1) \(\frac{x^2+2x^4}{4}\); 2) \(\frac{3y^5 - y^3}{3}\); 3) \(\frac{a-1}{a}\); 4) \(\frac{b^2+3b-3}{b^2}\).