Решебник по алгебре 8 класс Мерзляк ФГОС Задание 764

 

\[\boxed{\mathbf{764\ (764).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ y = \frac{x² - 2x - 8}{x - 4}\]

\[x^{2} - 2x - 8 = 0\]

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

\[x_{1} \cdot x_{2} = - 8,\ \ x_{2} = - 2\]

\[y = \frac{x^{2} - 2x - 8}{x - 4} =\]

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

\[y = x + 2,\ \ x \neq 4\]

\[2)\ y = \frac{x^{2} - x - 2}{x + 1} - \frac{x^{2} - x - 30}{x + 5}\]

\[x² - x - 2 = 0\]

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

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

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

\[= x - 2\]

\[x^{2} - x - 30 = 0\]

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

\[x_{1} \cdot x_{2} = - 30,\ \ x_{2} = - 5\]

\[\frac{x^{2} - x - 30}{x + 5} = \frac{(x - 6)(x + 5)}{(x + 5)} =\]

\[= x - 6\]

\[y = x - 2 - x + 6\]

\[y = 4,\ \ x \neq - 1,\ \ x \neq - 5\]

\[\boxed{\mathbf{7}\mathbf{6}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[2x_{1} - 5x_{2} = 28\]

\[x^{2} - 7x + m = 0\]

\[2 \cdot \left( 7 - x_{2} \right) - 5x_{2} = 28\]

\[14 - 2x_{2} - 5x_{2} = 28\]

\[- 7x_{2} = 14\]

\[x_{2} = - 2\]

\[x_{1} = 7 + 2 = 9\]

\[x_{1}x_{2} = m\]

\[- 2 \cdot 9 = - 18\]

\[Ответ:\ x_{1} = 9;\ \ \ x_{2} = - 2;\ \ \ \]

\[m = - 18.\]