Решебник по алгебре 9 класс Макарычев Задание 297

 

\[\boxed{\text{297}\text{\ (297)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{12}{x^{2} - 2x + 3} = x^{2} - 2x - 1\]

\[Пусть\ t = x^{2} - 2x - 1;\ \]

\[\ t + 4 = x^{2} - 2x + 3:\]

\[\frac{12}{t + 4} = t^{\backslash t + 4};\ \ \ \ \ t \neq - 4\ \ \ \ \ \ \ \ \]

\[t^{2} + 4t - 12 = 0\]

\[D_{1} = 4 + 12 = 16\]

\[t_{1,2} = - 2 \pm 4 = - 6;2.\]

\[\left\{ \begin{matrix} x^{2} - 2x - 1 = - 6 \\ x^{2} - 2x - 1 = 2\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 2x + 5 = 0\ \ (1) \\ x^{2} - 2x - 3 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ D = 1 - 5 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет;\]

\[(2)D = 1 + 3 = 4\ \]

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

\[Ответ:x = 4;\ \ x = - 1.\]

\[\textbf{б)}\ \frac{12}{x^{2} + x - 10} - \frac{6}{x^{2} + x - 6} =\]

\[= \frac{5}{x^{2} + x - 11}\]

\[Пусть\ x^{2} + x - 11 = a:\]

\[\ \frac{12}{a + 1} - \frac{6}{a + 5} = \frac{5}{a}\text{\ \ \ \ \ \ }\]

\[12a(a + 5) - 6a(a + 1) = a\]

\[a^{2} + 24a - 25 = 0\]

\[D_{1} = 144 + 25 = 169\]

\[a_{1} = - 12 + 13 = 1;\ \ \ a_{2} =\]

\[= - 12 - 13 = - 25.\]

\[\left\{ \begin{matrix} x^{2} + x - 11 = 1\ \ \ \ \ \ \\ x^{2} + x - 11 = - 25 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + x - 12 = 0\ \ (1) \\ x^{2} + x + 14 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ x^{2} + x - 12 = 0\]

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

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

\[(2)\ \ D = 1 - 14 = - 13 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:x = 3;\ \ x = - 4.\]

\[\textbf{в)}\ \frac{16}{x^{2} - 2x} - \frac{11}{x^{2} - 2x + 3} =\]

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

\[Пусть\ \ x^{2} - 2x + 1 = a:\]

\[\frac{16^{\backslash a(a + 2)}}{a - 1} - \frac{11^{\backslash a(a - 1)}}{a + 2} =\]

\[= \frac{9^{\backslash\text{(}a - 1)(a + 2)}}{a}\]

\[16a(a + 2) - 11a(a - 1) =\]

\[= 9 \cdot (a - 1)(a + 2)\]

\[4a^{2} - 34a - 18 = 0\ \ \ \ \ |\ :2\]

\[2a^{2} - 17a - 9 = 0\]

\[D = 17^{2} + 4 \cdot 2 \cdot 9 = 361\]

\[a_{1} = \frac{17 - 19}{4} = - \frac{1}{2};\ \]

\[\ a_{2} = \frac{17 + 19}{4} = 9.\]

\[\left\{ \begin{matrix} (x - 1)^{2} = 9\ \ \ \ \ \ \ \ \ \ \ (1) \\ (x - 1)^{2} = - \frac{1}{2}\ \ \ \ \ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\text{\ \ }(x - 1)^{2} = 9\ \ \]

\[x - 1 = 3;\ \ \ \ x - 1 = - 3\]

\[x = 4;\ \ \ \ \ \ \ \ \ \ \ \ x = - 2.\]

\[(2)\ (x - 1)^{2} = - \frac{1}{2}\]

\[корней\ нет,\ так\ как\ (x - 1)^{2}\text{\ \ }\]

\[всегда\ больше\ или\ равно\ 0.\]

\[Ответ:x = 4;\ \ x = - 2.\]

\[\boxed{\text{297.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{x - 8}{x + 4} > 2\]

\[\frac{x - 8}{x + 4} - 2^{\backslash x + 4} > 0\]

\[\frac{x - 8 - 2x - 8}{x + 4} > 0\]

\[\frac{- x - 16}{x + 4} > 0\]

\[\frac{x + 16}{x + 4} < 0\]

\[(x + 16)(x + 4) < 0\]

\[x \in ( - 16;\ - 4).\]

\[\textbf{б)}\ \frac{3 - x}{x - 2} < 1\]

\[\frac{3 - x}{x - 2} - 1^{\backslash x - 2} < 0\]

\[\frac{3 - x - x + 2}{x - 2} < 0\]

\[\frac{- 2x + 5}{x - 2} < 0\]

\[\frac{2x - 5}{x - 2} > 0\]

\[2 \cdot (x - 2)(x - 2,5) > 0\]

\[x \in ( - \infty;2) \cup (2,5;\ + \infty).\]

\[\textbf{в)}\ \frac{7x - 1}{x} > 5\]

\[\frac{7x - 1}{x} - 5^{\backslash x} > 0\]

\[\frac{7x - 1 - 5x}{x} > 0\]

\[\frac{2x - 1}{x} > 0\]

\[2 \cdot x \cdot (x - 0,5) > 0\]

\[x \in ( - \infty;0) \cup (0,5; + \infty).\]

\[\textbf{г)}\ \frac{6 - 2x}{x + 4} > 3\]

\[\frac{6 - 2x}{x + 4} - 3^{\backslash x + 4} > 0\]

\[\frac{6 - 2x - 3x - 12}{x + 4} > 0\]

\[\frac{- 5x - 6}{x + 4} > 0\]

\[\frac{5x + 6}{x + 4} < 0\]

\[5 \cdot (x + 1,2)(x + 4) < 0\]

\[x \in ( - 4; - 1,2).\]