ГДЗ по алгебре 9 класс Макарычев Задание 244

 

\[\boxed{\text{244}\text{\ (244)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = 3x^{2} - 0,5x + \frac{1}{16} \Longrightarrow\]

\[\Longrightarrow парабола,\ ветви\ \]

\[направлены\ вверх.\]

\[x_{b} = - \frac{b}{2a} = - \frac{- 0,5}{2 \cdot 3} = \frac{0,5}{6} = \frac{1}{12};\]

\[y_{b} = y\left( \frac{1}{12} \right) = 3 \cdot \frac{1}{144} - \frac{1}{2} \cdot \frac{1}{12} +\]

\[+ \frac{1}{16} = \frac{1}{48} - \frac{1}{24} + \frac{1}{16} = \frac{2}{48} =\]

\[= \frac{1}{24} \Longrightarrow\]

\[\Longrightarrow E(y) = \left\lbrack \frac{1}{24}; + \infty \right);\]

\[\textbf{б)}\ y = 2x^{2} + 1,2x + 2 \Longrightarrow\]

\[\Longrightarrow парабола,\ ветви\ \]

\[направлены\ вверх.\]

\[x_{b} = - \frac{b}{2a} = - \frac{1,2}{4} = - 0,3;\]

\[y_{b} = y( - 0,3) =\]

\[= 2 \cdot 0,09 - 1,2 \cdot 0,3 + 2 =\]

\[= 1,82 \Longrightarrow\]

\[\Longrightarrow E(y) = \lbrack 1,82; + \infty);\]

\[\textbf{в)}\ y = - \frac{1}{2}x^{2} + 4x - 5,5 \Longrightarrow\]

\[\Longrightarrow парабола,\ ветви\ \]

\[направлены\ вниз.\]

\[x_{b} = - \frac{b}{2a} = - \frac{4}{2 \cdot \left( - \frac{1}{2} \right)} =\]

\[= - \frac{4}{- 1} = 4;\]

\[y_{b} = y(4) = - \frac{1}{2} \cdot 16 + 4 \cdot 4 -\]

\[- 5,5 = - 8 + 16 - 5,5 = 2,5 \Longrightarrow\]

\[\Longrightarrow E(y) = ( - \infty;2,5\rbrack;\]

\[\textbf{г)}\ y = - 3x^{2} - 2x - 4\frac{2}{3} \Longrightarrow\]

\[\Longrightarrow парабола,\ ветви\ \]

\[направлены\ вниз.\]

\[x_{b} = - \frac{b}{2a} = - \frac{- 2}{2 \cdot ( - 3)} = - \frac{1}{3};\]

\[y_{b} = y\left( - \frac{1}{3} \right) =\]

\[= - 3 \cdot \frac{1}{9} + 2 \cdot \frac{1}{3} - 4\frac{2}{3} = - 4\frac{1}{3} \Longrightarrow\]

\[\Longrightarrow E(y) = \left( - \infty;\ - 4\frac{1}{3} \right\rbrack.\]

\[\boxed{\text{244.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[Пусть\ x - это\ искомое\ число.\ \]

\[Тогда\ 13 \cdot \left( x + \frac{1}{x} \right) = x^{3} + \frac{1}{x^{3}}.\]

\[Пусть\ \ x + \frac{1}{x} = t,\ \ t > 0,\]

\[\ \ тогда:\]

\[x^{3} + \frac{1}{x^{3}} = \left( x + \frac{1}{x} \right)\left( x^{2} - 1 + \frac{1}{x^{2}} \right)\]

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

\[= t^{2} - 2\]

\[x^{3} + \frac{1}{x^{3}} = t \cdot \left( t^{2} - 2 - 1 \right) =\]

\[= t \cdot \left( t^{2} - 3 \right).\]

\[\Longrightarrow 13t = t \cdot \left( t^{2} - 3 \right)\]

\[13t = t^{3} - 3t\]

\[t³ - 16t = 0\]

\[t\left( t^{2} - 16 \right) = 0\]

\[t(t - 4)(t + 4) = 0\]

\[t_{1} = 0,\ \ t_{2,3} = \pm 4.\]

\[Так\ как\ t > 0,\ то\ \ \]

\[\ x + \frac{1}{x} = 4\]

\[x^{2} - 4x + 1 = 0,\]

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

\[Ответ:x = 2 \pm \sqrt{3}.\]