Решение задания 73

1. a) $$\frac{2}{y} + \frac{3}{x} = \frac{2x}{xy} + \frac{3y}{xy} = \frac{2x + 3y}{xy}$$
2. б) $$\frac{a}{b^2} - \frac{b}{a} = \frac{a^2}{ab^2} - \frac{b^3}{ab^2} = \frac{a^2 - b^3}{ab^2}$$
3. г) $$\frac{2x}{3} - \frac{3x}{2} = \frac{4x}{6} - \frac{9x}{6} = \frac{4x - 9x}{6} = \frac{-5x}{6}$$
4. д) $$\frac{5x}{x} + \frac{8y}{4y} = 5 + 2 = 7$$
5. ж) $$1 - \frac{8}{5a} - \frac{25a}{8} = \frac{40a}{40a} - \frac{64}{40a} - \frac{125a^2}{40a} = \frac{40a - 64 - 125a^2}{40a}$$
6. з) $$\frac{3b}{c} + \frac{4c}{3b} + \frac{2b}{c} = \frac{3b}{c} + \frac{2b}{c} + \frac{4c}{3b} = \frac{9b}{3c} + \frac{6b}{3c} + \frac{4c}{3b} = \frac{15b}{3c} + \frac{4c}{3b} = \frac{5b}{c} + \frac{4c}{3b} = \frac{15b^2}{3bc} + \frac{4c^2}{3bc} = \frac{15b^2 + 4c^2}{3bc}$$