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МатематикаРешебник по алгебре 9 класс Макарычев Задание 242
Задание 242
\[\boxed{\text{242\ (242).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\boxed{\text{242.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \frac{12}{x^{2} - 2x + 3} = x² - 2x - 1\]
\[Пусть\ t = x^{2} - 2x - 1,\]
\[\ \ t + 4 = x^{2} - 2x + 3,\]
\[\frac{12}{t + 4} = t,\ \ t^{2} + 4t - 12 = 0,\ \]
\[\ t \neq - 4.\]
\[D = 4 + 12 = 16\]
\[t_{1,2} = - 2 \pm 4 = - 6;2.\]
\[\left\{ \begin{matrix} x^{2} - 2x - 1 = - 6 \\ x^{2} - 2x - 1 = 2\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 2x + 5 = 0\ \ (1) \\ x^{2} - 2x - 3 = 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ D = 1 - 5 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет;\]
\[(2)D = 1 + 3 = 4\ \]
\[\ x_{1,2} = 1 \pm 2 = 3;\ - 1.\]
\[Ответ:x = 4;\ \ x = - 1.\]
\[\textbf{б)}\ \frac{12}{x^{2} + x - 10} - \frac{6}{x^{2} + x - 6} =\]
\[= \frac{5}{x^{2} + x - 11}\]
\[Пусть\ x^{2} + x - 11 = a,\ тогда\]
\[\ \frac{12}{a + 1} - \frac{6}{a + 5} = \frac{5}{a};\]
\[12a(a + 5) - 6a(a + 1) = a,\ \ \]
\[a^{2} + 24a - 25 = 0;\]
\[a_{1} = 1,\ \ a_{2} = - 25 \Longrightarrow\]
\[\left\{ \begin{matrix} x^{2} + x - 11 = 1\ \ \ \ \ \\ x^{2} + x - 11 = - 25 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x² + x - 12 = 0\ \ (1) \\ x² + x + 14 = 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\text{\ \ }x_{1} = 3,\ \ x_{2} = - 4;\]
\[(2)\ \ D = 1 - 14 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет.\]
\[Ответ:x = 3;\ \ x = - 4.\]
\[\textbf{в)}\ \frac{16}{x^{2} - 2x} - \frac{11}{x^{2} - 2x + 3} =\]
\[= \frac{9}{x^{2} - 2x + 1}\]
\[Пусть\ \ x^{2} - 2x + 1 = a \Longrightarrow\]
\[\Longrightarrow \frac{16}{a - 1} - \frac{11}{a + 2} = \frac{9}{a};\]
\[16a(a + 2) - 11a(a - 1) =\]
\[= 9 \cdot (a - 1)(a + 2)\]
\[4a^{2} - 34a - 18 = 0\]
\[2a^{2} - 17a - 9 = 0\]
\[D = 17^{2} + 4 \cdot 2 \cdot 9 = 361\]
\[a_{1,2} = \frac{17 \pm 19}{4} = - \frac{1}{2};\ \ 9.\]
\[\left\{ \begin{matrix} (x - 1)^{2} = 9\ \ \ \ \ \ \ \ \ (1) \\ (x - 1)^{2} = - \frac{1}{2}\ \ \ \ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ \ \ x - 1 = \pm 3,\ \ \]
\[x_{1,2} = 4; - 2.\]
\[(2)\ корней\ нет.\]
\[Ответ:x = 4;\ \ x = - 2.\]
