Решение заданий на сложение и вычитание дробей.

a) $$\frac{x}{y^2} + \frac{2}{y} = \frac{x}{y^2} + \frac{2y}{y^2} = \frac{x+2y}{y^2}$$

б) $$\frac{2x}{y} - \frac{x}{y^2} = \frac{2xy}{y^2} - \frac{x}{y^2} = \frac{2xy - x}{y^2} = \frac{x(2y - 1)}{y^2}$$

в) $$\frac{1-a}{a^2} + \frac{1}{a} = \frac{1-a}{a^2} + \frac{a}{a^2} = \frac{1-a+a}{a^2} = \frac{1}{a^2}$$

г) $$\frac{x+a}{x^2} + \frac{x-a}{ax} = \frac{a(x+a)}{ax^2} + \frac{x(x-a)}{ax^2} = \frac{ax+a^2+x^2-ax}{ax^2} = \frac{x^2+a^2}{ax^2}$$

д) $$\frac{y+b}{y^2} + \frac{y+b}{by} = \frac{b(y+b)}{by^2} + \frac{y(y+b)}{by^2} = \frac{by+b^2+y^2+by}{by^2} = \frac{y^2+2by+b^2}{by^2} = \frac{(y+b)^2}{by^2}$$

e) $$\frac{a-2b}{ab^2} - \frac{b-2a}{a^2b} = \frac{a(a-2b)}{a^2b^2} - \frac{b(b-2a)}{a^2b^2} = \frac{a^2-2ab - (b^2-2ab)}{a^2b^2} = \frac{a^2-2ab - b^2+2ab}{a^2b^2} = \frac{a^2 - b^2}{a^2b^2}$$

ж) $$\frac{c-d}{c^2d} - \frac{c+d}{cd^2} = \frac{d(c-d)}{c^2d^2} - \frac{c(c+d)}{c^2d^2} = \frac{cd-d^2 - (c^2+cd)}{c^2d^2} = \frac{cd-d^2 - c^2-cd}{c^2d^2} = \frac{-d^2 - c^2}{c^2d^2} = -\frac{c^2+d^2}{c^2d^2}$$

з) $$\frac{a^2-b^2}{b^5} + \frac{2}{b^3} = \frac{a^2-b^2}{b^5} + \frac{2b^2}{b^5} = \frac{a^2-b^2+2b^2}{b^5} = \frac{a^2+b^2}{b^5}$$

и) $$\frac{x^3+1}{x^4} - \frac{1}{x} = \frac{x^3+1}{x^4} - \frac{x^3}{x^4} = \frac{x^3+1-x^3}{x^4} = \frac{1}{x^4}$$