Решение

в) $$\frac{a-b}{5} + \frac{4a-b}{10} = \frac{2(a-b) + 4a - b}{10} = \frac{2a - 2b + 4a - b}{10} = \frac{6a - 3b}{10} = \frac{3(2a - b)}{10}$$

е) $$\frac{m-n}{m^2} - \frac{n-m}{mn} = \frac{(m-n)n - (n-m)m}{m^2n} = \frac{mn - n^2 - nm + m^2}{m^2n} = \frac{m^2 - n^2}{m^2n} = \frac{(m-n)(m+n)}{m^2n}$$

б) $$\frac{3x+2}{5x} - \frac{5x+3y}{10xy} - \frac{y-1}{2y} = \frac{(3x+2)2y - (5x+3y) - (y-1)5x}{10xy} = \frac{6xy + 4y - 5x - 3y - 5xy + 5x}{10xy} = \frac{xy + y}{10xy} = \frac{y(x+1)}{10xy} = \frac{x+1}{10x}$$

в) $$\frac{4a}{3a-6} + \frac{3a}{8-4a} = \frac{4a}{3(a-2)} - \frac{3a}{4(a-2)} = \frac{16a - 9a}{12(a-2)} = \frac{7a}{12(a-2)}$$

в) $$\frac{4}{c^2-9} - \frac{2}{c^2+3c} = \frac{4}{(c-3)(c+3)} - \frac{2}{c(c+3)} = \frac{4c - 2(c-3)}{c(c-3)(c+3)} = \frac{4c - 2c + 6}{c(c-3)(c+3)} = \frac{2c + 6}{c(c-3)(c+3)} = \frac{2(c+3)}{c(c-3)(c+3)} = \frac{2}{c(c-3)}$$