Выполни сложение или вычитание дробей.

Конечно, давай выполним сложение и вычитание дробей из твоего задания. Вот решения: 1. а) $\frac{x}{3} + \frac{x-2}{5} = \frac{5x + 3(x-2)}{15} = \frac{5x + 3x - 6}{15} = \frac{8x - 6}{15}$ б) $\frac{3y-2}{6} - \frac{y+1}{4} = \frac{2(3y-2) - 3(y+1)}{12} = \frac{6y - 4 - 3y - 3}{12} = \frac{3y - 7}{12}$ в) $\frac{-b}{7} + \frac{3b}{14} = \frac{-2b + 3b}{14} = \frac{b}{14}$ г) $\frac{1}{a^2} + \frac{a-2}{a} = \frac{1 + a(a-2)}{a^2} = \frac{1 + a^2 - 2a}{a^2} = \frac{a^2 - 2a + 1}{a^2} = \frac{(a-1)^2}{a^2}$ д) $\frac{3x}{y} - \frac{5}{x} - \frac{3y-2}{y} = \frac{3x^2 - 5y - x(3y-2)}{xy} = \frac{3x^2 - 5y - 3xy + 2x}{xy}$ с) $\frac{b-a}{ab} - \frac{a-b}{b^2} = \frac{b(b-a) - a(a-b)}{ab^2} = \frac{b^2 - ab - a^2 + ab}{ab^2} = \frac{b^2 - a^2}{ab^2}$ 2. а) $\frac{(x+y)^2}{6y} + \frac{(x-y)^2}{12y} - \frac{x^2-y^2}{4y} = \frac{2(x^2+2xy+y^2) + (x^2-2xy+y^2) - 3(x^2-y^2)}{12y} = \frac{2x^2+4xy+2y^2 + x^2-2xy+y^2 - 3x^2+3y^2}{12y} = \frac{2xy + 6y^2}{12y} = \frac{2y(x+3y)}{12y} = \frac{x+3y}{6}$ б) $\frac{3a+1}{7a} - \frac{7a+b}{14ab} = \frac{2b(3a+1) - (7a+b)}{14ab} = \frac{6ab+2b - 7a - b}{14ab} = \frac{6ab + b - 7a}{14ab}$ 3. а) $\frac{a-1}{2(a-4)} + \frac{a}{a-4} = \frac{a-1 + 2a}{2(a-4)} = \frac{3a-1}{2(a-4)}$ б) $\frac{x-1}{3x-12} - \frac{x-3}{2x-8} = \frac{x-1}{3(x-4)} - \frac{x-3}{2(x-4)} = \frac{2(x-1) - 3(x-3)}{6(x-4)} = \frac{2x-2 - 3x+9}{6(x-4)} = \frac{-x+7}{6(x-4)}$ в) $\frac{3y}{4y-4} + \frac{2y}{3-3y} = \frac{3y}{4(y-1)} - \frac{2y}{3(y-1)} = \frac{9y - 8y}{12(y-1)} = \frac{y}{12(y-1)}$ 4. а) $\frac{a+1}{a^2 - ab} - \frac{1-6}{b^2 - ab} = \frac{a+1}{a(a-b)} - \frac{-5}{b(b-a)} = \frac{a+1}{a(a-b)} + \frac{5}{b(a-b)} = \frac{b(a+1) + 5a}{ab(a-b)} = \frac{ab+b+5a}{ab(a-b)}$ б) $\frac{3x^2-8y^2}{x^2-2xy} - \frac{3xy-x^2}{xy-2y^2} = \frac{3x^2-8y^2}{x(x-2y)} - \frac{3xy-x^2}{y(x-2y)} = \frac{y(3x^2-8y^2) - x(3xy-x^2)}{xy(x-2y)} = \frac{3x^2y - 8y^3 - 3x^2y + x^3}{xy(x-2y)} = \frac{x^3 - 8y^3}{xy(x-2y)}$ в) $\frac{2}{y^2-4} - \frac{1}{y^2+2y} = \frac{2}{(y-2)(y+2)} - \frac{1}{y(y+2)} = \frac{2y - (y-2)}{y(y-2)(y+2)} = \frac{2y - y + 2}{y(y-2)(y+2)} = \frac{y+2}{y(y-2)(y+2)} = \frac{1}{y(y-2)}$