Решебник по алгебре 8 класс Мерзляк Задание 783

\[\boxed{\mathbf{783\ (783).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\frac{x^{2} - 9x + 18}{x^{2} - 9} = 0;\ \ \ \ \ \ x \neq 3;\ \ \ \ \]

\[x \neq - 3\]

\[x^{2} - 9x + 18 = 0\]

\[x_{1} + x_{2} = 9,\ \ x_{1}x_{2} = 18,\ \ \]

\[x_{1} = 6,\ \ \]

\[x_{2} = 3\ (не\ подходит)\]

\[Ответ:x = 6.\]

\[2)\ \frac{3x^{2} - 14x - 5}{3x^{2} + x} = 0;\ \ \ \ \ \ \ \ \ \]

\[x \neq 0;\ \ \ \ \ x \neq - \frac{1}{3}\]

\[3x^{2} - 14x - 5 = 0\]

\[x_{1} + x_{2} = \frac{14}{3},\ \ x_{1}x_{2} = \frac{5}{3},\ \ \]

\[x_{1} = - \frac{1}{3}\ (не\ подходит),\ \ \]

\[x_{2} = \frac{15}{3} = 5\]

\[Ответ:x = 5.\]

\[3)\ \frac{x^{2} - 12x + 35}{x^{2} - 10x + 25} = 0\]

\[x^{2} - 12x + 35 = 0\]

\[x_{1} + x_{2} = 12,\ \ x_{1}x_{2} = 35,\ \ \]

\[x_{1} = 7,\ \ \]

\[x_{2} = 5\ (не\ подходит)\]

\[\frac{(x - 7)(x - 5)}{(x - 5)^{2}} = 0;\ \ \ \ \ \ x \neq 5\]

\[Ответ:x = 7.\]

\[4)\ \frac{x^{2} - 7x + 6}{x^{2} + 2x - 3} = 0\]

\[x^{2} - 7x + 6 = 0\]

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

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

\[x^{2} + 2x - 3 = 0\]

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

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

\[\frac{(x - 6)(x - 1)}{(x + 3)(x - 1)} = 0\]

\[\left\{ \begin{matrix} x - 6 = 0 \\ x - 1 \neq 0 \\ x + 3 \neq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 6 \\ x \neq 1 \\ x \neq - 3 \\ \end{matrix} \right.\ \]

\[Ответ:x = 6.\]