Решебник по алгебре 8 класс Мерзляк ФГОС Задание 751

 

\[\boxed{\mathbf{751\ (751).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x² - x - 12 = 0\]

\[x_{1} + x_{2} = 1,\ \ x_{1} = 4\]

\[x_{1} \cdot x_{2} = - 12,\ \ x_{2} = - 3\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[2)\ x² + 2x - 35 = 0\]

\[x_{1} + x_{2} = - 2,\ \ x_{1} = - 7\]

\[x_{1} \cdot x_{2} = - 35,\ \ x_{2} = 5\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[3)\ 3x² - 16x + 5 = 0\]

\[D = 256 - 4 \cdot 3 \cdot 5 = 196\]

\[x = \frac{16 \pm 14}{6}\]

\[x_{1} = 5;\ \ \ \ x_{2} = \frac{1}{3}\]

\[Ответ:\ x = 5;\ \ x = \frac{1}{3}.\]

\[4)\ 16x² - 24x + 3 = 0\]

\[D = 576 - 4 \cdot 3 \cdot 16 = 384\]

\[x = \frac{24 \pm 8\sqrt{6}}{32} = \frac{3 \pm \sqrt{6}}{4}\]

\[x_{1} = \frac{3 + \sqrt{6}}{4};\ \ \ \ \ \ \ \ x_{2} = \frac{3 - \sqrt{6}}{4}\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[5)\ 4x² + 28x + 49 = 0\]

\[D = 784 - 4 \cdot 4 \cdot 49 = 0\]

\[x = - \frac{28}{8} = - 3,5\]

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[6)\ 3x² + 21x - 90 = 0\]

\[D = 441 + 4 \cdot 3 \cdot 90 = 1521\]

\[x = \frac{- 21 \pm 39}{6}\]

\(x_{1} = - 10\); \(x_{2} = 3\)

\[x_{1} + x_{2} = - 5;\ \ \ \ x_{1} \cdot x_{2} = - 16\]

\[\boxed{\mathbf{7}\mathbf{5}\mathbf{1}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x_{1} = - 7\]

\[x_{2} = - 8\]

\[x_{1} + x_{2} = - 15\]

\[x_{1} \cdot x_{2} = 56\]

\[Уравнение:\ x² + 15x + 56 = 0\ \]

\[2)\ x_{1} = 5\]

\[x_{2} = - 0,4\]

\[x_{1} + x_{2} = 4,6\]

\[x_{1} \cdot x_{2} = - 2\ \]

\[x² - 4,6x - 2 = 0\ \ \ \ | \cdot 5\]

\[Уравнение:\ 5x^{2} - 23x - 10 = 0\]

\[3)\ x_{1} = \frac{1}{2}\]

\[x_{2} = \frac{2}{3}\]

\[x_{1} + x_{2} = \frac{7}{6}\]

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

\[x² - \frac{7}{6}x + \frac{2}{6} = 0\ \ | \cdot 6\]

\[Уравнение:\ 6x² - 7x + 2 = 0\]

\[4)\ x_{1} = 5 - \sqrt{10}\]

\[x_{2} = 5 + \sqrt{10}\]

\[x_{1} + x_{2} = 10\]

\[x_{1} \cdot x_{2} = 25 - 10 = 15\ \]

\[Урвнение:\ x² - 10x + 15 = 0\]