Решебник по алгебре 9 класс Мерзляк Задание 982

\[\boxed{\mathbf{982\ (982).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 5x² - x + a > 0\]

\[D = 1 - 20a < 0\]

\[1 - 20a < 0,\ \ 20a > 1,\]

\[\ \ a > \frac{1}{20}\]

\[Ответ:a \in \left( \frac{1}{20}; + \infty \right).\]

\[2)\ ax² - 10x - 5 < 0,\ \ a < 0\]

\[D = 100 + 20a < 0\]

\[20a < - 100,\ \ a < - 5\]

\[Ответ:a \in ( - \infty;\ - 5).\]

\[3)\ ax² - 2 \cdot (a - 1)x + 4a \leq 0\]

\[\left\{ \begin{matrix} a < 0 \\ D \leq 0 \\ \end{matrix} \right.\ \]

\[D = 4 \cdot (a - 1)^{2} - 16a^{2} =\]

\[= 4 \cdot \left( a^{2} - 2a + 1 \right) - 16a^{2} =\]

\[= 4a^{2} - 8a + 4 - 16a^{2} =\]

\[= - 12a^{2} - 8a + 4 \leq 0\ \ \ |\ :( - 4)\]

\[3a^{2} - 2a - 1 \geq 0\]

\[D = 4 + 12 = 16\]

\[a = \frac{- 2 + 4}{6} = \frac{1}{3},\ \ \]

\[a = \frac{- 2 - 4}{6} = - 1\]

\[Ответ:a \in ( - \infty; - 1\rbrack.\]

\[4)\ (a - 1)x² - (a + 1)x +\]

\[+ a + 1 > 0\]

\[\left\{ \begin{matrix} a - 1 > 0 \\ D < 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a > 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (a + 1)^{2} - 4 \cdot (a + 1)(a - 1) < 0 \\ \end{matrix} \right.\ \]

\[a^{2} + 2a + 1 - 4 \cdot \left( a^{2} - 1 \right) < 0\]

\[a^{2} + 2a + 1 - 4a^{2} + 4 < 0\]

\[- 3a^{2} + 2a + 5 < 0\ \ \ \ | \cdot ( - 1)\]

\[3a^{2} - 2a - 5 > 0\]

\[D = 4 + 60 = 64\]

\[a = \frac{2 + 8}{6} = \frac{5}{3},\]

\[\ \ a = \frac{2 - 8}{6} = - 1\]

\[Ответ:a \in \left( 1\frac{2}{3};\ + \infty \right).\]