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### Решение задания 20: 1) $\frac{3a}{10} + \frac{2a}{10} = \frac{3a+2a}{10} = \frac{5a}{10} = \frac{a}{2}$ 2) $\frac{6x}{5y} - \frac{x}{5y} = \frac{6x-x}{5y} = \frac{5x}{5y} = \frac{x}{y}$ 3) $\frac{2m-4n}{21c} + \frac{5m+18n}{21c} = \frac{2m-4n+5m+18n}{21c} = \frac{7m+14n}{21c} = \frac{7(m+2n)}{21c} = \frac{m+2n}{3c}$ 4) $\frac{2a+5b}{ab} - \frac{2a-3b}{ab} = \frac{2a+5b-(2a-3b)}{ab} = \frac{2a+5b-2a+3b}{ab} = \frac{8b}{ab} = \frac{8}{a}$ 5) $\frac{5y}{y^2-9} - \frac{15}{y^2-9} = \frac{5y-15}{y^2-9} = \frac{5(y-3)}{(y-3)(y+3)} = \frac{5}{y+3}$ 6) $\frac{y^2+8y}{4-y^2} - \frac{4y-4}{4-y^2} = \frac{y^2+8y-(4y-4)}{4-y^2} = \frac{y^2+8y-4y+4}{4-y^2} = \frac{y^2+4y+4}{4-y^2} = \frac{(y+2)^2}{(2-y)(2+y)} = \frac{y+2}{2-y} = -\frac{y+2}{y-2}$ ### Решение задания 21: 1) $\frac{x-4}{x-2} - \frac{x}{2-x} = \frac{x-4}{x-2} + \frac{x}{x-2} = \frac{x-4+x}{x-2} = \frac{2x-4}{x-2} = \frac{2(x-2)}{x-2} = 2$ 2) $\frac{5x+6}{5-x} + \frac{3x+16}{x-5} = \frac{-(5x+6)}{x-5} + \frac{3x+16}{x-5} = \frac{-5x-6+3x+16}{x-5} = \frac{-2x+10}{x-5} = \frac{-2(x-5)}{x-5} = -2$ 3) $\frac{(2a-1)^2}{6a-6} + \frac{(a-2)^2}{6-6a} = \frac{(2a-1)^2}{6(a-1)} - \frac{(a-2)^2}{6(a-1)} = \frac{(2a-1)^2-(a-2)^2}{6(a-1)} = \frac{(2a-1-a+2)(2a-1+a-2)}{6(a-1)} = \frac{(a+1)(3a-3)}{6(a-1)} = \frac{3(a+1)(a-1)}{6(a-1)} = \frac{a+1}{2}$ 4) $\frac{16-7x}{(x-4)^2} - \frac{x-x^2}{(4-x)^2} = \frac{16-7x}{(x-4)^2} - \frac{x-x^2}{(x-4)^2} = \frac{16-7x-(x-x^2)}{(x-4)^2} = \frac{16-7x-x+x^2}{(x-4)^2} = \frac{x^2-8x+16}{(x-4)^2} = \frac{(x-4)^2}{(x-4)^2} = 1$