Решение №3

1. a) $$\frac{7a-2b}{a-b} - \frac{8b-3a}{a-b} = \frac{7a-2b - (8b-3a)}{a-b} = \frac{7a-2b - 8b + 3a}{a-b} = \frac{10a - 10b}{a-b} = \frac{10(a-b)}{a-b} = 10$$
2. б) $$\frac{6x}{1-3x} + 2 = \frac{6x + 2(1-3x)}{1-3x} = \frac{6x + 2 - 6x}{1-3x} = \frac{2}{1-3x}$$
3. в) $$\frac{m-3}{m^2} + \frac{3m+1}{m^2} = \frac{m-3 + 3m+1}{m^2} = \frac{4m - 2}{m^2} = \frac{2(2m - 1)}{m^2}$$
4. г) $$\frac{9}{x+3} - \frac{3x}{x+3} = \frac{9 - 3x}{x+3} = \frac{3(3 - x)}{x+3}$$
5. д) $$\frac{9a^3}{b^3} : \frac{3a^9}{b^6} = \frac{9a^3}{b^3} \cdot \frac{b^6}{3a^9} = \frac{9a^3b^6}{3a^9b^3} = \frac{3b^3}{a^6}$$
6. e) $$\frac{4b^3}{b-2} \cdot \frac{b^2-4b+4}{8b^2} = \frac{4b^3}{b-2} \cdot \frac{(b-2)^2}{8b^2} = \frac{4b^3(b-2)^2}{8b^2(b-2)} = \frac{b(b-2)}{2} = \frac{b^2 - 2b}{2}$$