в) 1/(2a^7) + (4-2a^3)/a^10;

в) $\frac{1}{2a^7} + \frac{4-2a^3}{a^{10}} = \frac{a^3 + 2(4-2a^3)}{2a^{10}} = \frac{a^3+8-4a^3}{2a^{10}} = \frac{8-3a^3}{2a^{10}}$ г) $\frac{a+b}{a^2} + \frac{a-b}{ab} = \frac{b(a+b) + a(a-b)}{a^2b} = \frac{ab+b^2+a^2-ab}{a^2b} = \frac{a^2+b^2}{a^2b}$ б) $\frac{3x+5}{35x} + \frac{x-3}{21x} = \frac{3(3x+5) + 5(x-3)}{105x} = \frac{9x+15+5x-15}{105x} = \frac{14x}{105x} = \frac{14}{105} = \frac{2}{15}$ а) $\frac{15a-b}{12a} - \frac{a-4b}{9a} = \frac{3(15a-b) - 4(a-4b)}{36a} = \frac{45a-3b-4a+16b}{36a} = \frac{41a+13b}{36a}$ б) $\frac{a+3}{a^2-1} - \frac{1}{a^2+a} = \frac{a+3}{(a-1)(a+1)} - \frac{1}{a(a+1)} = \frac{a(a+3) - 1(a-1)}{a(a-1)(a+1)} = \frac{a^2+3a-a+1}{a(a^2-1)} = \frac{a^2+2a+1}{a(a^2-1)} = \frac{(a+1)^2}{a(a-1)(a+1)} = \frac{a+1}{a(a-1)} = \frac{a+1}{a^2-a}$ в) $\frac{x-12a}{x^2-16a^2} - \frac{4a}{4ax-x^2} = \frac{x-12a}{(x-4a)(x+4a)} - \frac{4a}{x(4a-x)} = \frac{x-12a}{(x-4a)(x+4a)} + \frac{4a}{x(x-4a)} = \frac{x(x-12a) + 4a(x+4a)}{x(x-4a)(x+4a)} = \frac{x^2-12ax+4ax+16a^2}{x(x^2-16a^2)} = \frac{x^2-8ax+16a^2}{x(x^2-16a^2)} = \frac{(x-4a)^2}{x(x-4a)(x+4a)} = \frac{x-4a}{x^2+4ax}$ г) $\frac{a-30y}{a^2-100y^2} - \frac{10y}{10ay-a^2} = \frac{a-30y}{(a-10y)(a+10y)} + \frac{10y}{a(a-10y)} = \frac{a(a-30y) + 10y(a+10y)}{a(a-10y)(a+10y)} = \frac{a^2-30ay+10ay+100y^2}{a(a^2-100y^2)} = \frac{a^2-20ay+100y^2}{a(a^2-100y^2)} = \frac{(a-10y)^2}{a(a-10y)(a+10y)} = \frac{a-10y}{a^2+10ay}$ в) $\frac{(a+b)^2}{a^2+ab} + \frac{(a-b)^2}{a^2-ab} = \frac{(a+b)^2}{a(a+b)} + \frac{(a-b)^2}{a(a-b)} = \frac{a+b}{a} + \frac{a-b}{a} = \frac{a+b+a-b}{a} = \frac{2a}{a} = 2$ г) $\frac{x^2-4}{5x-10} - \frac{x^2+4x+4}{5x+10} = \frac{(x-2)(x+2)}{5(x-2)} - \frac{(x+2)^2}{5(x+2)} = \frac{x+2}{5} - \frac{x+2}{5} = 0$