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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 245
Задание 245
\[\boxed{\text{245}\text{\ (245)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[h = 24t - 4,9t^{2} \Longrightarrow парабола,\ \]
\[ветви\ направлены\ вниз.\]
\[Наибольшая\ высота,\ которую\ \]
\[достиг\ мяч - ордината\ \]
\[вершины\ \]
\[параболы.\]
\[t_{b} = - \frac{b}{2a} = - \frac{24}{2 \cdot ( - 4,9)} = \frac{12}{4,9} =\]
\[= \frac{120}{49} = 2\frac{22}{49};\]
\[h_{b} = h\left( \frac{120}{4} \right) = 24 \cdot \frac{120}{4} -\]
\[- 4,9 \cdot \frac{120^{2}}{49^{2}} = \frac{24 \cdot 120}{49} -\]
\[- \frac{49 \cdot 120^{2}}{10 \cdot 49^{2}} =\]
\[= \frac{24 \cdot 120}{49} - \frac{12 \cdot 120}{49} =\]
\[= 120 \cdot \left( \frac{24 - 12}{49} \right) =\]
\[= 120 \cdot \frac{12}{49} = \frac{1440}{49} = 29\frac{19}{49}\ (м).\]
\[h(t) \Longrightarrow 24t - 4,9t^{2} = 0,\ \ \]
\[t(24 - 4,9t) = 0;\]
\[t_{1} = 0\ \ \ и\ \ t_{2} = \frac{24}{4,9} = \frac{240}{49} =\]
\[= 4\frac{44}{49}\ (с).\]
\[Мяч\ поднимался\ при\ \]
\[t \in \left\lbrack 0;2\frac{22}{49} \right\rbrack.\]
\[Мяч\ опускался\ \ при\ \]
\[t \in \left\lbrack 2\frac{22}{49};4\frac{44}{49} \right\rbrack.\]
\[Через\ 4\frac{44}{49}\ с\ мяч\ упал\ на\ землю.\]
\[\boxed{\text{245.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ x² + \frac{1}{x²} - \frac{1}{2} \cdot \left( x - \frac{1}{x} \right) = 3\frac{1}{2}\]
\[Пусть\ \ t = x - \frac{1}{x},\ \ \]
\[t^{2} = \left( x - \frac{1}{x} \right)^{2} = x^{2} + \frac{1}{x^{2}} - 2 \Longrightarrow\]
\[\Longrightarrow t^{2} + 2 - \frac{1}{2}t = \frac{7}{2};\]
\[2t^{2} + 4 - t - 7 = 0\]
\[2t^{2} - t - 3 = 0\]
\[D = 1 + 4 \cdot 2 \cdot 3 = 25\]
\[t_{1,2} = \frac{1 \pm 5}{4} = \frac{3}{2};\ - 1;\]
\[1)\ при\ t_{1} = \frac{3}{2} \Longrightarrow x - \frac{1}{x} = \frac{3}{2},\ \]
\[x^{2} - 1,5x - 1 = 0\]
\[2x^{2} - 3x - 2 = 0\]
\[D = 9 + 16 = 25\]
\[x_{1,2} = \frac{3 \pm 5}{4} = 2;\ - 0,5.\]
\[2)\ при\ t_{2} = - 1 \Longrightarrow x - \frac{1}{x} = - 1,\]
\[x^{2} + x - 1 = 0\]
\[D = 1 + 4 = 5\]
\[x_{1,2} = \frac{- 1 \pm \sqrt{5}}{2}.\]
\[Ответ:\ \ 2;\ - 0,5;\ \frac{- 1 \pm \sqrt{5}}{2}.\]
\[\textbf{б)}\ x² + \frac{1}{x^{2}} - \frac{1}{3} \cdot \left( x + \frac{1}{x} \right) = 8\]
\[Пусть\ t = x + \frac{1}{x},\]
\[\text{\ \ }t^{2} = \left( x + \frac{1}{x} \right)^{2} =\]
\[= x^{2} + \frac{1}{x^{2}} - 2 \Longrightarrow\]
\[\Longrightarrow t^{2} - 2 - \frac{1}{3}t = 8,\]
\[\ \ 3t^{2} - t - 30 = 0,\]
\[t_{1,2} = - 3;\ \frac{10}{3}.\]
\[1)\ при\ t_{1} = - 3 \Longrightarrow x + \frac{1}{x} = - 3,\]
\[x² + 3x + 1 = 0\]
\[D = 9 - 4 = 5\]
\[x_{1.2} = \frac{- 3 \pm \sqrt{5}}{2};\ \]
\[2)\ при\ t_{2} = \frac{10}{3} \Longrightarrow x + \frac{1}{x} = \frac{10}{3},\ \]
\[3x^{2} - 10x + 3 = 0\]
\[x_{1} = 3,\ \ x_{2} = \frac{1}{3}.\]
\[Ответ:3;\frac{1}{3};\ \frac{- 3 \pm \sqrt{5}}{2}\text{.\ }\]
