ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1018

\[\boxed{\text{1018.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ a > 3\]

\[\left( \frac{a - 3}{a + 3} - \frac{a + 3}{a - 3} \right) \cdot \left( 1 + \frac{3}{a} \right) < 0\]

\[\frac{(a - 3)^{2} - (a + 3)^{2}}{a^{2} - 9} \cdot \frac{a + 3}{a} < 0\]

\[\frac{a^{2} - 6a + 9 - a^{2} - 6a - 9}{a(a - 3)} < 0\]

\[\frac{- 12a}{a(a - 3)} < 0\]

\[\frac{- 12}{a - 3} < 0\]

\[при\ a > 3\ \ неравенство\ \]

\[\frac{- 12}{a - 3} < 0\ \ верно \Longrightarrow ч.т.д.\]

\[\textbf{б)}\ y > 1\]

\[\frac{y^{2} + 3}{y - 1} - \frac{2}{y}\ :\]

\[\ :\left( \frac{1}{y^{2} - y} + \frac{y - 3}{y^{2} - 1} \right) > 0\]

\[\frac{y^{2} + 3}{y - 1} - \frac{2}{y}\ :\]

\[\ :\left( \frac{1}{y(y - 1)} + \frac{y - 3}{(y - 1)(y + 1)} \right) > 0\]

\[\frac{y^{2} + 3}{y - 1} - \frac{2}{y}\ :\]

\[\ :\left( \frac{y + 1 + y^{2} - 3y}{y(y - 1)(y + 1)} \right) > 0\]

\[\frac{y^{2} + 3}{y - 1} - \frac{2}{y} \cdot \frac{y(y - 1)(y + 1)}{y^{2} - 2y + 1} > 0\]

\[\frac{y^{2} + 3}{y - 1} - \frac{2 \cdot (y - 1)(y + 1)}{(y - 1)^{2}} > 0\]

\[\frac{y^{2} + 3 - 2 \cdot (y + 1)}{y - 1} > 0\]

\[\frac{y^{2} + 3 - 2y - 2}{y - 1} > 0\]

\[\frac{y^{2} - 2y + 1}{y - 1} > 0\]

\[\frac{(y - 1)^{2}}{y - 1} > 0\]

\[y - 1 > 0\]

\[при\ y > 1\ \ неравенство\]

\[\ y - 1 > 0\ верно \Longrightarrow ч.т.д.\]