ЗАДАНИЯ №6 ОГЭ ПО МАТЕМАТИКЕ ДЕЙСТВИЯ С ОБЫКНОВЕННЫМИ ДРОБЯМИ 1) Выполните действия: a) $$\frac{x}{y^2} - \frac{1}{x} \div (\frac{1}{y} - \frac{1}{x});$$ r) $$\left(\frac{y}{y-5} - 2y\right) \div \frac{11-2y}{y-5};$$ б) $$(2+\frac{m}{m+1})\cdot \frac{3m^2+3m}{12m+8};$$ д) $$\frac{a+8b}{2b} \cdot \frac{3a^2}{b^2} \cdot \frac{b}{6a};$$ в) $$\frac{4+b}{4-b} \div (\frac{2b^2}{4+b} - b);$$ 2) a) $$\frac{x^2-4}{9-y^2} \div \frac{x-2}{3+y} \cdot \frac{2}{3-y};$$ б) $$\frac{a+b}{3a-b} + \frac{1}{a+b} - \frac{a^2-b^2}{3a-b};$$ в) $$\frac{1}{x-1} \div \frac{x+1}{x^2+x+1} \div (1+\frac{1}{x^3-1});$$ 2. Представьте в виде дроби: a) $$\left(\frac{m-4}{m+1} + \frac{m+4}{m-4}\right) \cdot \frac{m^2-16}{16};$$ б) $$\left(\frac{7}{b+7} + \frac{b^2+49}{b^2-49} - \frac{7}{b-7}\right) \div \frac{b+1}{2}.$$

Выполним действия. 1. a) $$\frac{x}{y^2} - \frac{1}{x} \div (\frac{1}{y} - \frac{1}{x}) = \frac{x}{y^2} - \frac{1}{x} \div (\frac{x-y}{xy}) = \frac{x}{y^2} - \frac{1}{x} \cdot \frac{xy}{x-y} = \frac{x}{y^2} - \frac{y}{x-y} = \frac{x(x-y) - y^3}{y^2(x-y)} = \frac{x^2 - xy - y^3}{y^2(x-y)}$$ г) $$\left(\frac{y}{y-5} - 2y\right) \div \frac{11-2y}{y-5} = \frac{y - 2y(y-5)}{y-5} \div \frac{11-2y}{y-5} = \frac{y - 2y^2 + 10y}{y-5} \cdot \frac{y-5}{11-2y} = \frac{11y - 2y^2}{11-2y} = \frac{y(11-2y)}{11-2y} = y$$ б) $$(2+\frac{m}{m+1})\cdot \frac{3m^2+3m}{12m+8} = \frac{2(m+1)+m}{m+1} \cdot \frac{3m(m+1)}{4(3m+2)} = \frac{3m+2}{m+1} \cdot \frac{3m(m+1)}{4(3m+2)} = \frac{3m}{4}$$ д) $$\frac{a+8b}{2b} \cdot \frac{3a^2}{b^2} \cdot \frac{b}{6a} = \frac{(a+8b)3a^2b}{2b \cdot b^2 \cdot 6a} = \frac{(a+8b)3a^2b}{12ab^3} = \frac{(a+8b)a}{4b^2} = \frac{a(a+8b)}{4b^2}$$ в) $$\frac{4+b}{4-b} \div (\frac{2b^2}{4+b} - b) = \frac{4+b}{4-b} \div \frac{2b^2 - b(4+b)}{4+b} = \frac{4+b}{4-b} \div \frac{2b^2 - 4b - b^2}{4+b} = \frac{4+b}{4-b} \div \frac{b^2 - 4b}{4+b} = \frac{4+b}{4-b} \cdot \frac{4+b}{b^2 - 4b} = \frac{(4+b)^2}{(4-b)b(b-4)} = -\frac{(4+b)^2}{b(4-b)^2}$$ 2. a) $$\frac{x^2-4}{9-y^2} \div \frac{x-2}{3+y} \cdot \frac{2}{3-y} = \frac{(x-2)(x+2)}{(3-y)(3+y)} \cdot \frac{3+y}{x-2} \cdot \frac{2}{3-y} = \frac{(x+2)(3+y)2}{(3-y)(3+y)(x-2)} = \frac{2(x+2)}{(3-y)^2}$$ б) $$\frac{a+b}{3a-b} + \frac{1}{a+b} - \frac{a^2-b^2}{3a-b} = \frac{(a+b)^2 + (3a-b) - (a^2-b^2)(a+b)}{(3a-b)(a+b)} = \frac{a^2+2ab+b^2 + 3a - b - (a-b)(a+b)^2}{(3a-b)(a+b)} = \frac{a^2+2ab+b^2 + 3a - b - (a-b)(a^2+2ab+b^2)}{(3a-b)(a+b)} = \frac{a^2+2ab+b^2 + 3a - b - (a^3+2a^2b+ab^2 - a^2b-2ab^2-b^3)}{(3a-b)(a+b)} = \frac{a^2+2ab+b^2 + 3a - b - a^3-2a^2b-ab^2 + a^2b+2ab^2+b^3}{(3a-b)(a+b)} = \frac{-a^3 - a^2b + ab^2 + a^2 + 2ab + 3a + b^2 - b + b^3}{(3a-b)(a+b)}$$ в) $$\frac{1}{x-1} \div \frac{x+1}{x^2+x+1} \div (1+\frac{1}{x^3-1}) = \frac{1}{x-1} \cdot \frac{x^2+x+1}{x+1} \div (\frac{x^3-1+1}{x^3-1}) = \frac{1}{x-1} \cdot \frac{x^2+x+1}{x+1} \div \frac{x^3}{x^3-1} = \frac{1}{x-1} \cdot \frac{x^2+x+1}{x+1} \cdot \frac{x^3-1}{x^3} = \frac{1}{x-1} \cdot \frac{x^2+x+1}{x+1} \cdot \frac{(x-1)(x^2+x+1)}{x^3} = \frac{(x^2+x+1)(x-1)(x^2+x+1)}{(x-1)(x+1)x^3} = \frac{(x^2+x+1)^2}{(x+1)x^3}$$ 2. a) $$\left(\frac{m-4}{m+1} + \frac{m+4}{m-4}\right) \cdot \frac{m^2-16}{16} = \frac{(m-4)^2+(m+4)(m+1)}{(m+1)(m-4)} \cdot \frac{m^2-16}{16} = \frac{m^2-8m+16+m^2+5m+4}{(m+1)(m-4)} \cdot \frac{(m-4)(m+4)}{16} = \frac{2m^2-3m+20}{(m+1)(m-4)} \cdot \frac{(m-4)(m+4)}{16} = \frac{(2m^2-3m+20)(m+4)}{16(m+1)}$$ б) $$\left(\frac{7}{b+7} + \frac{b^2+49}{b^2-49} - \frac{7}{b-7}\right) \div \frac{b+1}{2} = \left(\frac{7(b-7) + b^2+49 - 7(b+7)}{(b+7)(b-7)}\right) \div \frac{b+1}{2} = \frac{7b-49+b^2+49-7b-49}{(b+7)(b-7)} \div \frac{b+1}{2} = \frac{b^2-49}{(b+7)(b-7)} \div \frac{b+1}{2} = \frac{b^2-49}{(b+7)(b-7)} \cdot \frac{2}{b+1} = \frac{(b-7)(b+7)}{(b+7)(b-7)} \cdot \frac{2}{b+1} = \frac{2}{b+1}$$