Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 389

\[\boxed{\mathbf{389}.}\]

\[1)\ \frac{3x^{2}}{x - 1} - \frac{7}{x + 1} = \frac{5x^{2} + 9}{x^{2} - 1};\ \ \]

\[\text{\ x} \neq \pm 1\]

\[3x^{2}(x + 1) - 7(x - 1) =\]

\[= 5x^{2} + 9\]

\[3x^{3} + 3x^{2} - 7x + 7 -\]

\[- 5x^{2} - 9 = 0\]

\[3x^{3} - 2x^{2} - 7x - 2 = 0\]

\[Делители:\ \pm 1;\ \pm 2.\]

\[Ответ:x = - \frac{1}{3};\ \ x = 2.\]

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

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

\[3x^{2} - 7x - 6 =\]

\[= - 3 \cdot (x - 3)\left( x + \frac{2}{3} \right) =\]

\[= - (x - 3)(3x + 2)\]

\[D = 49 + 72 = 121\]

\[x_{1} = \frac{7 + 11}{6} = 3;\ \ \ \]

\[x_{2} = \frac{7 - 11}{6} = - \frac{2}{3}.\]

\[ОДЗ:\ \ x \neq 3;\ \ x \neq - \frac{2}{3}.\]

\[\frac{1 - x}{x - 3} - \frac{2x}{3x + 2} =\]

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

\[(1 - x)(3x + 2) -\]

\[- 2x(x - 3) = - 4\]

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

\[- 2x^{2} + 6x + 4 = 0\]

\[- 5x^{2} + 7x + 6 = 0\]

\[5x^{2} - 7x - 6 = 0\]

\[D = 49 + 120 = 169\]

\[x_{1} = \frac{7 + 13}{10} = 2;\ \ \]

\[x_{2} = \frac{7 - 13}{6} = - \frac{6}{10} = - 0,6.\]

\[Ответ:x = - 0,6;x = 2.\]