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МатематикаРешебник по алгебре 9 класс Макарычев Задание 334
Задание 334
\[\boxed{\text{334\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x - 5}{x + 6} < 0\]
\[(x + 6)(x - 5) < 0\]
\[x \in ( - 6;5).\]
\[\textbf{б)}\ \frac{1,4 - x}{x + 3,8} < 0\]
\[(1,4 - x)(x + 3,8) < 0\]
\[(x + 3,8)(x - 1,4) > 0\]
\[x \in ( - \infty;\ - 3,8) \cup (1,4;\ + \infty).\]
\[\textbf{в)}\ \frac{2x}{x - 1,6} > 0\]
\[2x(x - 1,6) > 0\]
\[x \in ( - \infty;0) \cup (1,6;\ + \infty).\]
\[\textbf{г)}\ \frac{5x - 1,5}{x - 4} > 0\]
\[5 \cdot (x - 0,3)(x - 4) > 0\]
\[x \in ( - \infty;0,3) \cup (4;\ + \infty).\]
\[\textbf{д)}\ \frac{5x + 1}{x - 2} > 0\]
\[(5x + 1)(x - 2) > 0\]
\[5 \cdot (x + 0,2)(x - 2) > 0\]
\[x \in ( - \infty; - 0,2) \cup (2; + \infty).\]
\[\textbf{е)}\ \frac{3x}{2x + 9} < 0\]
\[3x(2x + 9) < 0\]
\[3x \cdot 2 \cdot (x + 4,5) < 0\]
\[6x(x + 4,5) < 0\]
\[x \in ( - 4,5;0).\]
\[\boxed{\text{334\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x - 5}{x + 6} < 0\]
\[(x - 5)(x + 6) < 0\]
\[(x + 6)(x - 5) < 0\]
\[x \in ( - 6;5).\]
\[\textbf{б)}\ \frac{1,4 - x}{x + 3,8} < 0\]
\[(1,4 - x)(x + 3,8) < 0\]
\[(x + 3,8)(x - 1,4) > 0\]
\[x \in ( - \infty;\ - 3,8) \cup (1,4;\ + \infty).\]
\[\textbf{в)}\ \frac{2x}{x - 1,6} > 0\]
\[2x(x - 1,6) > 0\]
\[x \in ( - \infty;0) \cup (1,6;\ + \infty).\]
\[\textbf{г)}\ \frac{5x - 1,5}{x - 4} > 0\]
\[(5x - 1,5)(x - 4) > 0\]
\[5 \cdot (x - 0,3)(x - 4) > 0\]
\[x \in ( - \infty;0,3) \cup (4;\ + \infty).\]
\[\boxed{\text{334.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \frac{x^{4}\ }{x^{2} - 2} + \frac{1 - 4x^{2}}{2 - x^{2}} + 4 = 0\]
\[\frac{x^{4}}{x^{2} - 2} - \frac{1 - 4x^{2}}{x^{2} - 2} + 4 = 0\]
\[\frac{x^{4} - 1 + 4x^{2} + 4x^{2} - 8}{x^{2} - 2} = 0\]
\[ОДЗ:x^{2} - 2 \neq 0,\ \ x^{2} \neq 2,\]
\[\ \ x \neq \pm \sqrt{2}.\]
\[x^{4} + 8x^{2} - 9 = 0;\]
\[Пусть\ \ y = x^{2},\ \ y^{2} = x^{4},\]
\[\ \ y \geq 0.\]
\[y^{2} + 8y - 9 = 0,\ \ по\ теореме\ \]
\[Виета:\]
\[y_{1} = 1,\ \ y_{2} = - 9.\]
\[Так\ как\ y \geq 0,\ \ то\ y = 1;\]
\[\Longrightarrow x^{2} = 1,\ \ x = \pm 1.\]
\[Пусть\ \ y = x^{2},\ \ y^{2} = x^{4},\]
\[\ \ y \geq 0,\]
\[\frac{y + 3}{y + 1} + \frac{2}{y - 4} + \frac{10}{y^{2} - 3y - 4} = 0\]
\[\frac{y + 3}{y + 1} + \frac{2}{y - 4} + \frac{10}{(y - 4)(y + 4)} = 0\]
\[\frac{(y + 3)(y - 4) + 2 \cdot (y + 1) + 10}{(y - 4)(y + 1)} = 0\]
\[ОДЗ:\ \ \ y \neq 4;\ - 1;\]
\[y^{2} - 4y + 3y - 12 + 2y + 2 + 10 = 0\]
\[y^{2} + y = 0\]
\[y(y + 1) = 0\]
\[y_{1} = 0,\ \ y_{2} = - 1.\]
\[Так\ как\ y \geq 0 \Longrightarrow y = 0 \Longrightarrow\]
\[\Longrightarrow x^{2} = 0 \Longrightarrow x = 0.\]
\[Ответ:\ \ а) - 1;1;\ \ \ б)\ 0.\]
