ГДЗ по алгебре 9 класс Макарычев Задание 940

\[\boxed{\text{940\ (940).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x}{x - 3} - \frac{5}{x + 3} = \frac{18}{x^{2} - 9},\]

\[x \neq \pm 3,\]

\[\frac{x(x + 3) - 5 \cdot (x - 3) - 18}{(x - 3)(x + 3)} = 0\]

\[x^{2} + 3x - 5x + 15 - 18 = 0\]

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

\[D = 4 + 12 = 16\]

\[x_{1} = \frac{2 + 4}{2} = 3 \Longrightarrow не\ подходит.\]

\[x_{2} = \frac{2 - 4}{2} = - 1.\]

\[Ответ:x = - 1.\]

\[\textbf{б)}\ \frac{70}{x^{2} - 16} - \frac{17}{x - 4} = \frac{3x}{x + 4},\]

\[x \neq \pm 4,\]

\[\frac{70 - 17 \cdot (x + 4) - 3x(x - 4)}{(x - 4)(x + 4)} =\]

\[= 0\]

\[70 - 17x - 68 - 3x^{2} + 12x = 0\]

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

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{- 5 + 7}{6} = \frac{1}{3},\]

\[x_{2} = \frac{- 5 - 7}{6} = - 2.\]

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

\[\textbf{в)}\ \frac{3}{(2 - x)^{2}} - \frac{5}{(x + 2)^{2}} = \frac{14}{x^{2} - 4},\]

\[x \neq \pm 2,\]

\[- 16x^{2} + 32x + 48 = 0\ \ \ |\ :( - 16)\]

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

\[D = 4 + 12 = 16\]

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

\[x_{2} = \frac{2 - 4}{2} = - 1.\]

\[Ответ:x = 3;\ \ x = - 1.\]

\[\frac{2 \cdot 2x + x(2 + x) - 7 \cdot (2 - x)}{2 \cdot (2 - x)(2 + x) \cdot x} =\]

\[= 0\]

\[4x + 2x + x^{2} - 14 + 7x = 0\]

\[x^{2} + 13x - 14 = 0\]

\[D = 169 + 56 = 225\]

\[x_{1} = \frac{- 13 + 15}{2} = 1,\]

\[x_{2} = \frac{- 13 - 15}{2} = - 14.\]

\[Ответ:x = - 14;\ \ x = 1.\]

\[\textbf{д)}\ \frac{1}{x^{2} - 9} + \frac{1}{3x - x^{2}} = \frac{3}{2x + 6},\ \]

\[x \neq \pm 3,\]

\[\frac{2x - 2 \cdot (x + 3) - 3x(x - 3)}{(x - 3)(x + 3) \cdot x \cdot 2} = 0\]

\[2x - 2x - 6 - 3x^{2} + 9x = 0\]

\[3x² - 9x + 6 = 0\ \ |\ :3\]

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

\[D = 9 - 8 = 1\]

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

\[x_{2} = \frac{3 - 1}{2} = 1.\]

\[Ответ:x = 2;\ \ x = 1.\]

\[\textbf{е)}\ \frac{2}{1 - x^{2}} - \frac{1}{1 - x} + \frac{4}{(x + 1)^{2}} = 0,\]

\[x \neq \pm 1,\]

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

\[= 0\]

\[x^{2} + 4x - 5 = 0\]

\[D = 16 + 20 = 36\]

\[x_{1} = \frac{- 4 + 6}{2} = 1 \Longrightarrow не\ подходит,\]

\[x_{2} = \frac{- 4 - 6}{2} = - 5.\]

\[Ответ:x = - 5.\]

\[\textbf{ж)}\ \frac{2}{x^{2} + 5x} + \frac{3}{2x - 10} = \frac{15}{x^{2} - 25},\ \]

\[\ x \neq \pm 5,\]

\[4x - 20 + 3x^{2} + 15x - 30x = 0\]

\[3x^{2} - 11x - 20 = 0\]

\[D = 121 + 240 = 361\]

\[x_{1} = \frac{11 + 19}{6} = 5 \Longrightarrow не\ \]

\[подходит.\]

\[x_{2} = \frac{11 - 19}{6} = - \frac{8}{6} = - 1\frac{1}{3}.\]

\[Ответ:x = - 1\frac{1}{3}.\]

\[\textbf{з)}\ \frac{5}{2x + 6} - \frac{1}{6x^{2} - 18x} =\]

\[= \frac{29}{27 - 3x^{2}}\]

\[15x^{2} - 45x - x - 3 + 58x = 0\]

\[15x^{2} + 12x - 3 = 0\ \ \ |\ :3\]

\[5x^{2} + 4x - 1 = 0\]

\[D = 16 + 20 = 36\]

\[x_{1} = \frac{- 4 + 6}{10} = \frac{2}{10} = \frac{1}{5},\]

\[x_{2} = \frac{- 4 - 6}{10} = - 1.\]

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