ГДЗ по алгебре 8 класс Мерзляк Задание 787

Авторы:
Год:2023
Тип:учебник
Серия:Алгоритм успеха

Задание 787

\[\boxed{\mathbf{787\ (787).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{10}{x + 2} + \frac{9}{x} = 1\]

\[\frac{10}{x + 2} + \frac{9}{x} - 1 = 0\]

\[\frac{10x + 9x + 18 - x^{2} - 2x}{x(x + 2)} = 0;\ \ \ \ \]

\[x \neq 0;\ \ \ \ x \neq - 2\]

\[- x^{2} + 17x + 18 = 0\]

\[x^{2} - 17x - 18 = 0\]

\[x_{1} + x_{2} = 17;\ \ x_{1}x_{2} = - 18,\ \ \]

\[x_{1} = 18,\ \ x_{2} = - 1\]

\[Ответ:\ x = - 1;x = 18.\]

\[2)\ \frac{48}{14 - x} - \frac{48}{14 + x} = 1\]

\[\frac{48}{14 - x} - \frac{48}{14 + x} - 1 = 0;\ \ \ \ \ \]

\[x \neq 14;\ \ \ \ x \neq - 14\]

\[x^{2} + 96x - 196 = 0\]

\[x_{1} + x_{2} = - 96,\ \ \]

\[x_{1}x_{2} = - 196,\ \]

\[\ x_{1} = 2,\ \ x_{2} = - 98\]

\[Ответ:\ x = - 98;x = 2.\]

\[3)\ \frac{x - 1}{x + 2} + \frac{x}{x - 2} = \frac{8}{x^{2} - 4}\]

\[(x - 1)(x - 2) + x(x + 2) - 8 =\]

\[= 0\]

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

\[= 0\ \ \]

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

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{4} = 2\ (не\ подходит);\ \ \ \ \ \ \]

\[x_{2} = \frac{1 - 7}{4} = - \frac{6}{4} = - 1,5\]

\[Ответ:\ x = - 1,5.\]

\[4)\ \frac{x - 1}{x + 3} + \frac{x + 1}{x - 3} = \frac{2x + 18}{x² - 9}\]

\[\frac{x - 1}{x + 3} + \frac{x + 1}{x - 3} - \frac{2x + 18}{(x - 3)(x + 3)} =\]

\[= 0;\ \ \ \ \ \ \ x \neq 3;\ \ x \neq - 3\]

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

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

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

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

\[x_{2} = 3\ (не\ подходит)\]

\[Ответ:\ x = - 2.\]

\[5)\ \frac{4x - 10}{x - 1} + \frac{x + 6}{x + 1} = 4\]

\[\frac{4x - 10}{x - 1} + \frac{x + 6}{x + 1} - 4 = 0;\ \ \ \ \]

\[x \neq 1;\ \ \ \ x \neq - 1\]

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

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

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

\[Ответ:\ x = - 3;x = 4.\]

\[6)\ \frac{1}{x} - \frac{10}{x^{2} - 5x} = \frac{3 - x}{x - 5}\]

\[\frac{1}{x} - \frac{10}{x(x - 5)} - \frac{3 - x}{x - 5} = 0;\ \ \ \ \ \]

\[x \neq 0;\ \ \ x \neq 5\]

\[x - 5 - 10 - (3 - x) \cdot x = 0\]

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

\[x^{2} - 2x - 15 = 0\]

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

\[x_{1} = 5\ (не\ подходит),\ \ \]

\[x_{2} = - 3\]

\[Ответ:\ x = - 3.\]

\[7)\ \frac{4x}{x^{2} + 4x + 4} - \frac{x - 2}{x^{2} + 2x} = \frac{1}{x}\]

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

\[x \neq 0;\ \ \ x \neq - 2\]

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

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

\[2x(x - 2) = 0\]

\[x = 0\ (не\ подходит);\ \ \ x = 2\]

\[Ответ:x = 2.\]

\[x^{2} - 15x + 54 = 0\]

\[x_{1} + x_{2} = 15,\ \ x_{1}x_{2} = 54,\ \ \]

\[x_{1} = 9,\ \ \]

\[x_{2} = 6\ (не\ подходит)\]

\[Ответ:x = 9.\]

\[9)\ \frac{x}{x + 7} + \frac{x + 7}{x - 7} = \frac{63 - 5x}{x^{2} - 49}\]

\[2x^{2} + 12x - 14 = 0\ \ \ \ \ |\ :2\]

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

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

\[\ x_{1} = - 7\ (не\ подходит),\ \ \]

\[x_{2} = 1\]

\[Ответ:x = 1.\]

\[10)\ \frac{4}{x^{2} - 10x + 25} - \frac{1}{x + 5} =\]

\[= \frac{10}{x² - 25}\]

\[- x^{2} + 4x + 45 = 0\]

\[x^{2} - 4x - 45 = 0\]

\[x_{1} + x_{2} = 4;\ \ \ \ \ x_{1} \cdot x_{2} = 4 - 5\ \]

\[x_{1} = - 5\ (не\ подходит);\ \ \ \ \]

\[x_{2} = 9\]

\[Ответ:\ x = 9.\]

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