\( \dfrac{5a}{6}\)

\(\dfrac{a}{2}+\dfrac{a}{3} = \dfrac{\;a^{(3}}{2}+\dfrac{\;a^{(2}}{3}=\dfrac{3a}{6}+\dfrac{2a}{6}= \dfrac{5a}{6} \)

\( \dfrac{7x}{6} \)

\(\dfrac{x}{2}+\dfrac{2x}{3}=\dfrac{\;\;x^{(3}}{2}+\dfrac{\;\;2x^{(2}}{3}=\dfrac{3x}{6}+\dfrac{4x}{6}=\dfrac{7x}{6} \)

\( -\dfrac{x}{8} \)

\(\dfrac{x}{4}-\dfrac{3x}{8}=\dfrac{\;\;x^{(2}}{4}-\dfrac{\;\;3x^{(1}}{8}= \dfrac{2x}{8}-\dfrac{3x}{8}=\dfrac{2x-3x}{8}=\dfrac{-x}{8}=-\dfrac{x}{8} \)

\( -\dfrac{xy}{8} \)

\(\dfrac{xy}{4}-\dfrac{3xy}{8}=\dfrac{\;\;xy^{(2}}{4}-\dfrac{\;\;3xy^{(1}}{8}= \dfrac{2xy}{8}-\dfrac{3xy}{8}=\dfrac{2xy-3xy}{8}=\dfrac{-xy}{8}=-\dfrac{xy}{8} \)

\( \dfrac{2x+3y}{8} \)

\(\dfrac{x+y}{4}+\dfrac{y}{8}=\dfrac{\;\;x+y^{\;\;(2}}{4}+\dfrac{\;\;y^{\;\;(1}}{8}= \dfrac{2(x+y)}{8}+\dfrac{y}{8}= \dfrac{2x+2y}{8}+\dfrac{y}{8}= \dfrac{2x+2y+y}{8} = \dfrac{2x+3y}{8} \)

\( \dfrac{x+2y}{6} \)

\(\dfrac{x+y}{4}+\dfrac{y-x}{12}= \dfrac{\;\;x+y^{\;\;(3}}{4}+\dfrac{\;\;y-x^{\;\;(1}}{12}= \dfrac{3(x+y)}{12}+\dfrac{y-x}{12}= \dfrac{3x+3y}{12}+\dfrac{y-x}{12}= \)


\(= \dfrac{3x+3y+y-x}{12} = \dfrac{2x+4y}{12}= \dfrac{2(x+2y)}{12}=\dfrac{x+2y}{6}\)

\( \dfrac{13y-12x}{15} \)

\(\dfrac{x+y}{5}+\dfrac{2y-3x}{3}= \dfrac{\;\;x+y^{\;\;(3}}{5}+\dfrac{\;\;2y-3x^{\;\;(5}}{3}= \dfrac{3(x+y)}{15}+\dfrac{5(2y-3x)}{15}= \)


\( \dfrac{3(x+y)+5(2y-3x)}{15} = \dfrac{3x+3y+10y-15x}{15} = \dfrac{13y-12x}{15} \)

\( x \)

\(\dfrac{x+y}{2}-\dfrac{y-x}{2}= \dfrac{x+y-(y-x)}{2}=\dfrac{x+y-y+x}{2}= \dfrac{2x}{2}=x\)

\( 2x \)

\(\dfrac{4x+5y}{4}-\dfrac{5y-4x}{4}= \dfrac{4x+5y-(5y-4x)}{4}=\dfrac{4x+5y-5y+4x}{4}= \dfrac{8x}{4}=2x\)

\( \dfrac{2y}{3} \)

\(\dfrac{4x-5y}{6}-\dfrac{4x-9y}{6}=\dfrac{4x-5y-(4x-9y)}{6}=\dfrac{4x-5y-4x+9y}{6}= \dfrac{4y}{6}= \dfrac{2y}{3} \)

\( -y \)

\(\dfrac{5x-19y}{10}-\dfrac{5x-9y}{10}=\dfrac{5x-19y-(5x-9y)}{10}=\dfrac{5x-19y-5x+9y}{10}= \dfrac{-10y}{10}= -y \)

\( -\dfrac{y}{2} \)

\( \dfrac{y+x}{xy} \)

\( \dfrac{1}{x}+\dfrac{1}{y}= \dfrac{\;\;\;\;1^{\;\;(y}}{x}+\dfrac{\;\;\;\;1^{\;\;(x}}{y}=\dfrac{y}{xy}+\dfrac{x}{xy}= \dfrac{y+x}{xy} \)

\( \dfrac{y+1}{xy} \)

\( \dfrac{1}{x}+\dfrac{1}{xy}= \dfrac{\;\;\;\;1^{\;\;(y}}{x}+\dfrac{\;\;\;\;1^{\;\;(1}}{xy}=\dfrac{y}{xy}+\dfrac{1}{xy}= \dfrac{y+1}{xy} \)

\( \dfrac{2y-1}{xy} \)

\( \dfrac{2}{x}-\dfrac{1}{xy}= \dfrac{\;\;\;\;2^{\;\;(y}}{x}-\dfrac{\;\;\;\;1^{\;\;(1}}{xy}=\dfrac{2y}{xy}-\dfrac{1}{xy}= \dfrac{2y-1}{xy} \)

\( \dfrac{y^2+1}{xy} \)

\( \dfrac{y}{x}+\dfrac{1}{xy}= \dfrac{\;\;\;\;y^{\;\;(y}}{x}+\dfrac{\;\;\;\;1^{\;\;(1}}{xy}=\dfrac{y^2}{xy}+\dfrac{1}{xy}= \dfrac{y^2+1}{xy} \)

\( 0 \)

\( \dfrac{2}{x}+\dfrac{2}{xy}-\dfrac{2(y+1)}{xy}= \dfrac{\;\;\;\;2^{\;\;(y}}{x}+\dfrac{\;\;\;\;2^{\;\;(1}}{xy}-\dfrac{2(y+1)^{\;\;(1}}{xy}= \dfrac{2y+2-2(y+1)}{xy}= \)

\(\dfrac{2y+2-2y-2}{xy}=\dfrac{0}{xy}=0 \)

\( 0 \)

\( \dfrac{5}{a}-\dfrac{5}{ab}-\dfrac{5(b-1)}{ab}= \dfrac{\;\;\;\;5^{\;\;(b}}{a}-\dfrac{\;\;\;\;5^{\;\;(1}}{ab}-\dfrac{5(b-1)^{\;\;(1}}{ab}= \dfrac{5b-5-5(b-1)}{ab}= \)

\(\dfrac{5b-5-5b+5}{ab}=\dfrac{0}{ab}=0 \)

\( -\dfrac{10}{b} \)

\( \dfrac{x}{ab}-\dfrac{5}{b}-\dfrac{x+5a}{ab}= \dfrac{\;\;\;\;x^{\;\;(1}}{ab}-\dfrac{\;\;\;\;5^{\;\;(a}}{b}-\dfrac{x+5a^{\;\;(1}}{ab}= \dfrac{x-5a-x-5a}{ab}= \)

\(\dfrac{-10a}{ab}=-\dfrac{10}{b} \)

\( -\dfrac{2c}{x} \)

\( \dfrac{7c}{xy}-\dfrac{c}{x}-\dfrac{7c+cy}{xy} = \dfrac{\;\;\;\;7c^{\;\;(1}}{xy}-\dfrac{\;\;\;\;c^{\;\;(y}}{x}-\dfrac{7c+cy^{\;\;(1}}{xy}= \dfrac{7c-cy-7c-cy}{xy}= \)

\(\dfrac{-2cy}{xy}=-\dfrac{2c}{x} \)

\( 0 \)

\( \dfrac{2a}{x}-\dfrac{18a}{xy}-\dfrac{2ay-18a}{xy} = \dfrac{\;\;\;\;2a^{\;\;(y}}{x}-\dfrac{\;\;\;\;18a^{\;\;(1}}{xy}-\dfrac{2ay-18a^{\;\;(1}}{xy}= \)

\( \dfrac{2ay-18a-(2ay-18a)}{xy}= \dfrac{2ay-18a-2ay+18a}{xy}=\dfrac{0}{xy}=0 \)

\( \dfrac{12a}{xy} \)

\( \dfrac{2a}{x}+\dfrac{6a}{xy}-\dfrac{2ay-6a}{xy} = \dfrac{\;\;\;\;2a^{\;\;(y}}{x}+\dfrac{\;\;\;\;6a^{\;\;(1}}{xy}-\dfrac{2ay-6a^{\;\;(1}}{xy}= \)

\( \dfrac{2ay+6a-(2ay-6a)}{xy}= \dfrac{2ay+6a-2ay+6a}{xy}=\dfrac{12a}{xy} \)

\( \dfrac{9a+3}{x} \)

\( \dfrac{2a+1}{x}+\dfrac{7ay+2y}{xy} = \dfrac{2a+1}{x}+\dfrac{y(7a+2)}{xy}= \dfrac{2a+1}{x}+\dfrac{7a+2}{x}=\dfrac{2a+1+7a+2}{x}= \)

\(=\dfrac{9a+3}{x} \)

Другое, менее рациональное решение

\(\dfrac{2a+1}{x}+\dfrac{7ay+2y}{xy} = \dfrac{\;\;2a+1^{\;\;(y}}{x}+\dfrac{7ay+2y^{\;\;(1}}{xy} = \dfrac{2ay+y}{xy}+\dfrac{7ay+2y}{xy} = \)

\( \dfrac{2ay+y+7ay+2y}{xy} = \dfrac{9ay+3y}{xy}=\dfrac{3y(3a+1)}{xy}=\dfrac{3(3a+1)}{x}=\dfrac{9a+3}{x}\)

\( \dfrac{8a}{x} \)

\( \dfrac{\;\;5a+2^{\;\;(y}}{x}+\dfrac{3ay+15^{\;\;(1}}{xy}-\dfrac{2y+15^{\;\;(1}}{xy}= \dfrac{5ay+2y+3ay+15-2y-15}{xy}= \)

\(=\dfrac{8ay}{xy}=\dfrac{8a}{x} \)

Другое решение

\( \dfrac{5a+2}{x}+\dfrac{3ay+15}{xy}-\dfrac{2y+15}{xy} = \dfrac{5a+2}{x}+\dfrac{3ay+15-2y-15}{xy}=\)

\(= \dfrac{5a+2^{\;\;(y}}{x}+\dfrac{3ay-2y^{\;\;(1}}{xy}=\dfrac{5ay+2y+3ay-2y}{xy}= \dfrac{8ay}{xy}=\dfrac{8a}{x}\)

\( \dfrac{yz+xz+xy}{xyz} \)

\( \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} = \dfrac{\;\;\;\;\;1^{\;\;(yz}}{x}+\dfrac{\;\;\;\;\;1^{\;\;(xz}}{y}+\dfrac{\;\;\;\;\;1^{\;\;(xy}}{z}= \dfrac{yz+xz+xy}{xyz} \)

\( \dfrac{1}{xy} \)

\( \dfrac{\;\;\;\;\;1^{\;\;(yz}}{x}+ \dfrac{\;\;\;\;\;1^{\;\;(z}}{xy}+\dfrac{\;\;\;\;1^{\;\;(xy}}{z}-\dfrac{yz+xy^{\;\;(1}}{xyz}= \dfrac{yz+z+xy-yz-xy}{xyz}= \)

\( \dfrac{z}{xyz}=\dfrac{1}{xy} \)

\( \dfrac{5}{6a} \)

\( (\dfrac{\;\;\;\;a^{\;\;(3}}{2}+\dfrac{\;\;\;\;a^{\;\;(2}}{3})\cdot \dfrac{1}{a^2}=\dfrac{3a+2a}{6}\cdot \dfrac{1}{a^2} =\dfrac{5a}{6}\cdot \dfrac{1}{a^2}=\dfrac{5}{6a} \)

\( \dfrac{2a}{(a-b)(a+b)} \)

\( \dfrac{1}{a+b}+\dfrac{1}{a-b}= \dfrac{\;\;\;\;\;\;1^{\;\;(a-b}}{a+b}+\dfrac{\;\;\;\;\;\;\;1^{\;\;(a+b}}{a-b}= \dfrac{a-b+a+b}{(a-b)(a+b)} = \)

\(= \dfrac{2a}{(a-b)(a+b)} \)

\( 1 \)

\(1)\;\;\;\; \dfrac{1}{a+b}+\dfrac{1}{a-b}= \dfrac{\;\;\;\;\;\;1^{\;\;(a-b}}{a+b}+\dfrac{\;\;\;\;\;\;\;1^{\;\;(a+b}}{a-b}= \dfrac{a-b+a+b}{(a-b)(a+b)} = \)

\(= \dfrac{2a}{(a-b)(a+b)} \)


\( 2) \;\;\;\; \dfrac{2a}{(a-b)(a+b)}\cdot \dfrac{(a+b)(a-b)}{2a}=1 \)

\( -1 \)

\(1)\;\;\;\; \dfrac{1}{a+b}-\dfrac{1}{a-b}= \dfrac{\;\;\;\;\;\;1^{\;\;(a-b}}{a+b}-\dfrac{\;\;\;\;\;\;\;1^{\;\;(a+b}}{a-b}= \dfrac{a-b-a-b}{(a-b)(a+b)} = \)

\(= \dfrac{-2b}{(a-b)(a+b)} \)


\( 2) \;\;\;\; \dfrac{-2b}{(a-b)(a+b)}\cdot \dfrac{(a+b)(a-b)}{2b}=-1 \)

\( 1 \)

\(1)\;\;\;\; \dfrac{2}{a-b}-\dfrac{1}{a+b}= \dfrac{\;\;\;\;\;\;2^{\;\;(a+b}}{a+b}-\dfrac{\;\;\;\;\;\;\;1^{\;\;(a-b}}{a-b}= \dfrac{2a+2b-1\cdot(a-b)}{(a-b)(a+b)} = \)

\(= \dfrac{2a+2b-a+b}{(a-b)(a+b)}=\dfrac{a+3b}{(a-b)(a+b)} \)


\( 2) \;\;\;\; \dfrac{a+3b}{(a-b)(a+b)}\cdot \dfrac{(a+b)(a-b)}{a+3b}=1 \)

\( \dfrac{1}{a+b} \)

\( 1)\;\;\;\dfrac{a}{a-b}-\dfrac{b}{a+b}=\dfrac{\;\;\;a^{\;\;(a+b}}{a-b}-\dfrac{\;\;\;b^{\;\;(a-b}}{a+b}= \dfrac{a(a+b)-b(a-b)}{a-b}= \)

\( \dfrac{a^2+ab-ba+b^2}{(a-b)(a+b)}=\dfrac{a^2+b^2}{(a-b)(a+b)} \)


\( 2) \;\;\;\; \dfrac{a^2+b^2}{(a-b)(a+b)}\cdot \dfrac{a-b}{a^2+b^2}=\dfrac{1}{a+b} \)

\( \dfrac{1}{c} \)

\( 1)\;\;\;\dfrac{2c}{c+d}+\dfrac{d-c}{c}=\dfrac{\;\;\;2c^{\;\;(c}}{c+d}+\dfrac{d-c^{\;\;(c+d}}{c}= \dfrac{2c\cdot c+(d-c)(c+d)}{(c+d)c}= \)

\(=\dfrac{2c^2+dc+d^2-c^2-cd}{(c+d)c}=\dfrac{c^2+d^2}{(c+d)c} \)


\( 2) \;\;\;\; \dfrac{c^2+d^2}{(c+d)c} \cdot \dfrac{c+d}{c^2+d^2}=\dfrac{1}{c} \)

\( 1 \)

\( 1)\;\;\; 1-\dfrac{1}{a}=\dfrac{1}{1}-\dfrac{1}{a}=\dfrac{\;\;\;\;1^{\;\;(a}}{1}-\dfrac{\;\;\;\;1^{\;\;(1}}{a}= \dfrac{a-1}{a} \)


\( 2) \;\;\;\; \dfrac{a-1}{a}\cdot \dfrac{a}{a-1}=1 \)

\( \dfrac{1}{b} \)

\( 1)\;\;\; b+\dfrac{1}{b}=\dfrac{b}{1}+\dfrac{1}{b}=\dfrac{\;\;\;\;b^{\;\;(b}}{1}+\dfrac{\;\;\;\;1^{\;\;(1}}{b} =\dfrac{b^2+1}{b}\)


\( 2) \;\;\;\; \dfrac{b^2+1}{b} \cdot \dfrac{1}{b^2+1}=\dfrac{1}{b} \)

\( 1 \)

\( 1)\;\;\; \dfrac{1}{b}-\dfrac{1}{a}=\dfrac{\;\;\;\;1^{\;\;(a}}{b}-\dfrac{\;\;\;\;1^{\;\;(b}}{a}=\dfrac{a-b}{ba} \)


\( 2)\;\;\; \dfrac{a-b}{ba}\cdot \dfrac{ab}{a-b}=1 \)

\( \dfrac{5}{a} \)

\( 1)\;\;\; 1+\dfrac{1}{a}= \dfrac{1}{1}+\dfrac{1}{a}=\dfrac{\;\;\;\;1^{\;\;(a}}{1}+\dfrac{\;\;\;\;1^{\;\;(1}}{a}=\dfrac{a+1}{a} \)


\( 2)\;\;\; \dfrac{a+1}{a}\cdot \dfrac{5}{a+1}=\dfrac{5}{a} \)

\( \dfrac{a+1}{a-1} \)

\( 1)\;\;\; 1+\dfrac{1}{a}=\dfrac{\;\;\;\;1^{\;\;(a}}{1}+\dfrac{\;\;\;\;1^{\;\;(1}}{a}=\dfrac{a+1}{a} \)

\( 2)\;\;\; 1-\dfrac{1}{a}=\dfrac{\;\;\;\;1^{\;\;(a}}{1}-\dfrac{\;\;\;\;1^{\;\;(1}}{a}=\dfrac{a-1}{a} \)

\( 3)\;\;\; \dfrac{a+1}{a} : \dfrac{a-1}{a}= \dfrac{a+1}{a} \cdot \dfrac{a}{a-1}= \dfrac{a+1}{a-1} \)