Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 113

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Задание 113

\[\boxed{\mathbf{113}.}\]

\[1)\ \frac{x^{2} - 5x + 6}{x^{2} - 5x} = 0\]

\[ОДЗ:\]

\[x^{2} - 5x \neq 0\]

\[x(x - 5) \neq 0\]

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

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

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 2;\ \ \ x_{2} = 3.\]

\[Ответ:x = 2;\ \ x = 3.\]

\[2)\ \frac{- x^{2} - 2x + 15}{x^{2} + 4x} = 0\]

\[ОДЗ:\]

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

\[x(x + 4) \neq 0\]

\[x \neq 0;\ \ x \neq - 4.\]

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

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

\[D_{1} = 1 + 15 = 16\]

\[x_{1} = - 1 + 4 = 3;\ \ \]

\[x_{2} = - 1 - 4 = - 5.\]

\[Ответ:x = - 5;\ \ x = 3.\]

\[3)\ \ \frac{x^{2} - x - 12}{x^{2} - 9} = 0\]

\[ОДЗ:\]

\[x^{2} - 9 \neq 0\]

\[x^{2} \neq 9\]

\[x \neq \pm 3.\]

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

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

\[x_{1} = 4;\ \ \]

\[x_{2} = - 3\ (не\ подходит).\]

\[Ответ:x = 4.\]

\[4)\ \frac{3x^{2} + 8x - 3}{2x + 6} = 0\]

\[ОДЗ:\]

\[2x + 6 \neq 0\]

\[2x \neq - 6\]

\[x \neq - 3.\]

\[3x^{2} + 8x - 3 = 0\]

\[D_{1} = 16 + 9 = 25\]

\[x_{1} = \frac{- 4 + 5}{3} = \frac{1}{3};\ \]

\[\ x_{2} = \frac{- 4 - 5}{3} =\]

\[= - 3\ (не\ подходит)\]

\[Ответ:x = \frac{1}{3}.\]

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

\[\frac{x(x - 3) + 4 \cdot 2(x - 3) - 5 \cdot 2x}{2x(x - 3)} = 0\]

\[ОДЗ:\]

\[2x(x - 3) \neq 0\]

\[x \neq 0;\ \ \ x \neq 3.\]

\[x^{2} - 3x + 8x - 24 - 10x = 0\]

\[x^{2} + 5x - 24 = 0\]

\[x_{1} + x_{2} = - 5;\ \ \ x_{1} \cdot x_{2} = - 24\]

\[x_{1} = - 8;\ \ \ \]

\[x_{2} = 3\ (не\ подходит).\]

\[Ответ:x = - 8.\]

\[6)\ \frac{7}{x} + \frac{1}{x - 5} = 1\frac{1}{2}\]

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

\[\frac{7 \cdot 2 \cdot (x - 5) + 2x - 3x(x - 5)}{2x(x - 5)} = 0\]

\[ОДЗ:\]

\[2x(x - 5) \neq 0\]

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

\[14x - 70 + 2x - 3x^{2} + 15x = 0\]

\[- 3x^{2} + 31x - 70 = 0\]

\[3x^{2} - 31x + 70 = 0\]

\[D = 961 - 840 = 121\]

\[x_{1} = \frac{31 + 11}{6} = 7;\ \ \ \]

\[x_{2} = \frac{31 - 11}{6} = \frac{20}{6} = \frac{10}{3} = 3\frac{1}{3}.\]

\[Ответ:x = 3\frac{1}{3};\ \ x = 7.\]

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