Представьте в виде дроби (2a-b)/a*(a/(2a-b)+a/b)

Ответ:

\[\mathbf{\ }\frac{2a - b}{a} \cdot \left( \frac{a^{\backslash b}}{2a - b} + \frac{a^{\backslash 2a - b}}{b} \right) = \frac{2a - b}{a} \cdot\]

\[\cdot \frac{ab + 2a^{2} - ab}{b(2a - b)} =\]

\[= \frac{(2a - b) \cdot 2a²}{\text{ab}(2a - b)}\mathbf{=}\frac{2a}{b}\]

\[y = \frac{4}{x}\]

\[Область\ определения\ функции:x \neq 0.\ \]

\[y > 0\ \ при\ \ x > 0.\]

\[\frac{2y}{y + 3} + (y - 3)^{2} \cdot \left( \frac{2}{9 - 6y + y^{2}} + \frac{1}{9 - y^{2}} \right) =\]

\[= \frac{2y}{y + 3} + (y - 3)^{2} \cdot \left( \frac{2^{\backslash 3 + y}}{(3 - y)^{2}} + \frac{1^{\backslash 3 - y}\ }{(3 - y)(3 + y)} \right) =\]

\[= \frac{2y}{y + 3} + (y - 3)^{2} \cdot \left( \frac{2 \cdot (3 + y) + 3 - y}{(3 - y)^{2}(3 + y)} \right) =\]

\[= \frac{2y}{y + 3} + (y - 3)^{2} \cdot \frac{6 + 2y + 3 - y}{(3 - y)^{2} \cdot (3 + y)} =\]

\[= \frac{2y}{y + 3} + \frac{9y}{3 + y} =\]

\[= \frac{2y + 9 + y}{y + 3} = \frac{3y + 9}{y + 3} = \frac{3 \cdot (y + 3)}{y + 3} =\]

\[= 3 - не\ зависит\ от\ y.\]

\[\frac{3x}{1^{\backslash 10 - 5x} - \frac{6}{10 - 5x}} = \frac{3x}{\frac{10 - 5x - 6}{10 - 5x}} =\]

\[= \frac{3x}{\frac{4 - 5x}{10 - 5x}} = \frac{3x(10 - 5x)}{4 - 5x} = \frac{30x - 15x²}{4 - 5x}\]

\[4 - 5x \neq 0\ \ \ \ \ \ \ \ \ \ \ 10 - 5x \neq 0\]

\[5x \neq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5x \neq 10\]

\[x \neq 0,8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 2\]

\[Выражение\ имеет\ смысл\ при\ x \neq 0,8;\ \ \]

\[x \neq 2.\]

\[\ \frac{14p^{4}}{q^{6}} \cdot \frac{q^{5}}{56p^{4}} = \frac{14p^{4}q^{5}}{56p^{4}q^{6}} = \frac{1}{4q}\]