Решебник по алгебре 8 класс Мерзляк Задание 791

\[\boxed{\mathbf{791\ (791).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{2x - 10}{x³ + 1} + \frac{4}{x + 1} = \frac{5x - 1}{x² - x + 1}\]

\[- x^{2} - 6x - 5 = 0\]

\[x^{2} + 6x + 5 = 0\]

\[x_{1} + x_{2} = - 6,\ \ x_{1}x_{2} = 5,\ \ \]

\[x_{1} = - 5,\ \ \]

\[x_{2} = - 1\ (не\ подходит)\]

\[Ответ:\ x = - 5.\]

\[2)\ \frac{6}{x^{2} - 4x + 3} + \frac{5 - 2x}{x - 1} = \frac{3}{x - 3}\]

\[x^{2} - 4x + 3 = (x - 3)(x - 1)\]

\[x_{1} + x_{2} = 4,\ \ x_{1}x_{2} = 3,\ \ \]

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

\[- 2x^{2} + 8x - 6 = 0\ \ \ \ \ |\ :( - 2)\]

\[x^{2} - 4x + 3 \neq 0\]

\[Ответ:корней\ нет.\]

\[3)\ \frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{14}{x^{2} + 3x + 2}\]

\[x^{2} + 3x + 2 = (x + 2)(x + 1)\]

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

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

\[3x^{2} - 4x - 20 = 0\]

\[D = 16 + 240 = 256\]

\[x = \frac{4 - 16}{6} = - 2\ (не\ подходит)\]

\[x = \frac{4 + 16}{6} = \frac{10}{3} = 3\frac{1}{3}\]

\[Ответ:x = 3\frac{1}{3}.\]

\[4)\ \frac{x}{x^{2} - 4} - \frac{3x - 1}{x^{2} + x - 6} =\]

\[= \frac{2}{x² + 5x + 6}\]

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

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

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

\[x^{2} + 5x + 6 = (x + 3)(x + 2)\]

\[x_{1} + x_{2} = - 5,\ \ x_{1}x_{2} = 6,\ \ \]

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

\[x \neq 2;\ \ \ x \neq - 2;\ \ x \neq - 3\]

\[- 2x^{2} - 4x + 6 = 0\ \ \ \ |\ :( - 2)\]

\[x² + 2x - 3 = 0\]

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

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

\[x_{2} = 1\]

\[Ответ:x = 1.\]