ГДЗ по алгебре 8 класс Мерзляк Задание 694

\[\boxed{\mathbf{694\ (694).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x² - (2a - 5)x - 3a^{2} + 5a =\]

\[= 0\]

\[D =\]

\[= (2a - 5)^{2} - 4 \cdot \left( - 3a^{2} + 5a \right) =\]

\[= 16a^{2} - 40a + 25 = (4a + 5)^{2}\]

\[x = \frac{(2a - 5) \pm (4a + 5)}{2}\]

\[x_{1} = \frac{6a}{2} = 3a\]

\[x_{2} = \frac{2a - 5 - 4a - 5}{2} =\]

\[= \frac{- 2a - 10}{2} = - a - 5\]

\[Ответ:x = 3a;x = \ - a - 5.\]

\[2)\ x² + (3a - 4)x - 12a = 0\]

\[D = (3a - 4)^{2} + 48a =\]

\[= 9a^{2} - 24a + 16 + 48a =\]

\[= 9a^{2} + 24a + 16 =\]

\[= (3a + 4)^{2}\]

\[x = \frac{( - 3a + 4) \pm (3a + 4)}{2}\]

\[x_{1} = \frac{- 3a + 4 + 3a + 4}{2} = \frac{8}{2} = 4\]

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

\[= - 3a\]

\[Ответ:x = 4;\ x = - 3a.\]

\[3)\ ax² - (a + 1)x + 1 = 0\]

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

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

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

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

\[если\ a \neq 0:\ \ \ \ \ x_{1} = 1;\ \ x_{2} = \frac{1}{a}.\]

\[если\ a = 0:\ - x + 1 = 0 \Longrightarrow \ \ \]

\[\Longrightarrow x = 1.\]

\[Ответ:\ \ x = 1;\ \ \ x = \frac{1}{a}.\]