Решебник по алгебре 8 класс Мерзляк ФГОС Задание 13

\[\boxed{\text{13\ (13).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[5x - 15y = 1\]

\[5 \cdot (x - 3y) = 1\]

\[x - 3y = \frac{1}{5}\]

\[1)\ x - 3y = \frac{1}{5}\]

\[2)\ \frac{8}{2x - 6y} = \frac{8}{2 \cdot (x - 3y)} =\]

\[= \frac{8}{2 \cdot \frac{1}{5}} = \frac{8 \cdot 5}{2} = \frac{40}{2} = 20\]

\[3)\ \frac{18y - 6x}{9} = \frac{- 6 \cdot (x - 3y)}{9} =\]

\[= \frac{- 6 \cdot \frac{1}{5}}{9} = \frac{- 2}{3 \cdot 5} = \frac{- 2}{15}\]

\[4)\ \frac{1}{x^{2} - 6xy + 9y^{2}} = \frac{1}{(x - 3y)^{2}} =\]

\[= \frac{1}{\left( \frac{1}{5} \right)^{2}} = \frac{(5)^{2}}{1} = 25\]

\[\boxed{\text{13.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[5x - 15y = 1\]

\[5 \cdot (x - 3y) = 1\]

\[x - 3y = \frac{1}{5}\]

\[1)\ x - 3y = \frac{1}{5}\]

\[2)\ \frac{8}{2x - 6y} = \frac{8}{2 \cdot (x - 3y)} =\]

\[= \frac{8}{2 \cdot \frac{1}{5}} = \frac{8 \cdot 5}{2} = \frac{40}{2} = 20\]

\[3)\ \frac{18y - 6x}{9} = \frac{- 6 \cdot (x - 3y)}{9} =\]

\[= \frac{- 6 \cdot \frac{1}{5}}{9} = \frac{- 2}{3 \cdot 5} = \frac{- 2}{15}\]

\[4)\ \frac{1}{x^{2} - 6xy + 9y^{2}} = \frac{1}{(x - 3y)^{2}} =\]

\[= \frac{1}{\left( \frac{1}{5} \right)^{2}} = \frac{(5)^{2}}{1} = 25\]