Решебник по алгебре 8 класс Мерзляк Задание 693

\[\boxed{\mathbf{693\ (693).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x² + (3a + 1)x + 2a² + a = 0\]

\[D = (3a + 1)^{2} - 4 \cdot \left( 2a^{2} + a \right) =\]

\[= 9a^{2} + 6a + 1 - 8a^{2} - 4a =\]

\[= a^{2} + 2a + 1 = (a + 1)^{2}\]

\[x = \frac{( - 3a - 1) \pm (a + 1)}{2}\]

\[x_{1} = \frac{- 3a - 1 + a + 1}{2} = \frac{- 2a}{2} =\]

\[= - a\]

\[x_{2} = \frac{- 3a - 1 - a - 1}{2} =\]

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

\[Ответ:\ x = - a;x = - 2a - 1.\]

\[2)\ x² - (2a + 4)x + 8a = 0\]

\[D = (2a + 4)^{2} - 32a =\]

\[= 4a^{2} + 16a + 16 - 32a =\]

\[= 4a² - 16a + 16 = (2a - 4)²\]

\[x = \frac{(2a + 4) \pm (2a - 4)}{2}\]

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

\[x_{2} = \frac{2a + 4 - 2a + 4}{2} = \frac{8}{2} = 4\]

\[Ответ:x = 2a;x = 4.\]

\[3)\ a²x² - 24ax - 25 = 0\]

\[D = 576a^{2} + 100a^{2} = 676a^{2} =\]

\[= (26a)^{2}\]

\[x = \frac{24a \pm 26a}{2a^{2}}\]

\[если\ a = 0,\ то\ корней\ нет;\]

\[если\ a \neq 0:\]

\[\ x_{1} = \frac{50a}{2a^{2}} = \frac{25}{a};\ \ \]

\[x_{2} = - \frac{2a}{2a^{2}} = - \frac{1}{a}.\]

\[Ответ:при\ a = 0\ нет\ корней;\ \ \ \]

\[при\ a \neq 0:\ x = \frac{25}{a},\ x = - \frac{1}{a}.\ \]

\[D = 4 \cdot (a + 1)^{2} - 4 \cdot (6a - 3) =\]

\[= 4a^{2} + 8a + 4 - 24a + 12 =\]

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

\[x = \frac{(2a + 2) \pm (2a - 4)}{6 \cdot (2a - 1)},\ \ \]

\[2a - 1 \neq 0,\ \ a \neq \frac{1}{2}\]

\[x_{1} = \frac{4a - 2}{6 \cdot (2a - 1)} =\]

\[= \frac{2 \cdot (2a - 1)}{6 \cdot (2a - 1)} = \frac{1}{3}\]

\[x_{2} = \frac{6}{6} = 1\]

\[Ответ:если\ a = \frac{1}{2}:корней\ нет;\]

\[если\ a \neq \frac{1}{2}:\ x = \frac{1}{3},\ x = 1.\]