ГДЗ по алгебре 9 класс Макарычев Задание 367

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Год:2020-2021-2022
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Задание 367

\[\boxed{\text{367\ (367).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{1}{x^{2} - 6x + 8} - \frac{1}{x - 2} +\]

\[+ \frac{10}{x^{2} - 4} = 0\]

\[x^{2} - 6x + 8 = (x - 2)(x - 4)\]

\[D_{1} = 9 - 8 = 1\]

\[x_{1} = 3 + 1 = 4;\ \ \ x_{2} = 3 - 1 = 2.\]

\[\frac{1^{\backslash x + 2}}{(x - 4)(x - 2)} - \frac{1^{\text{(}x - 4)(x + 2)}}{x - 2} +\]

\[+ \frac{10^{\backslash x - 4}}{(x - 2)(x + 2)} = 0\]

\[\frac{x + 2 - (x - 4)(x + 2) + 10 \cdot (x - 4)}{(x - 2)(x + 2)(x - 4)} = 0\]

\[ОДЗ:\ \ \ \ x \neq 2,\ \ x \neq - 2,\ \ \]

\[x \neq 4.\]

\[x + 2 - \left( x^{2} - 2x - 8 \right) + 10x -\]

\[- 40 = 0\]

\[- x^{2} + 13x - 30 = 0\]

\[x^{2} - 13x + 30 = 0\]

\[По\ теореме\ Виета:\ \]

\[x_{1} + x_{2} = 13;\ \ x_{1} \cdot x_{2} = 30\]

\[\ x_{1} = 10;\ \ \ \ x_{2} = 3.\]

\[\textbf{б)}\ \frac{3}{x^{2} - x - 6} + \frac{3}{x + 2} = \frac{7}{x^{2} - 9}\]

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

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

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

\[\frac{3^{\backslash x + 3}}{(x - 3)(x + 2)} + \frac{3^{\backslash\text{(}x - 3)(x + 3)}}{x + 2} -\]

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

\[\frac{3 \cdot (x + 3) + 3 \cdot \left( x^{2} - 9 \right) - 7 \cdot (x + 2)}{(x + 2)(x - 3)(x + 3)} = 0\]

\[ОДЗ:\ \ \ x \neq - 2,\ \ x \neq 3,\ \ \]

\[x \neq - 3.\]

\[3x + 9 + 3x^{2} - 27 - 7x -\]

\[- 14 = 0\]

\[3x^{2} - 4x - 32 = 0\]

\[D_{1} = 4 + 3 \cdot 32 = 100\]

\[x_{1,2} = \frac{2 \pm 10}{3} = 4;\ - \frac{8}{3}.\]

\[Ответ:а)\ 3;10;\ \ б) - 2\frac{2}{3};4.\]

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