Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 898

\[\boxed{\mathbf{898}\mathbf{.}}\]

\[f(x) = x^{3} - 6\]

\[f^{'}(a) = \left( x^{3} \right)^{'} - (6)^{'} =\]

\[= 3x^{2} - 0 = 3a^{2}\]

\[f(a) = a^{3} - 6\]

\[y_{1} = a^{3} - 6 + 3a^{2} \bullet (x - a) =\]

\[= a^{3} - 6 + 3xa^{2} - 3a^{3} =\]

\[= 3xa^{2} - 2a^{3} - 6\]

\[y_{2} = 3xb^{2} - 2b^{3} - 6.\]

\[Точки\text{\ A\ }и\ C:\]

\[y_{1}(0) = 3 \bullet 0 \bullet a^{3} - 2a^{2} - 6 =\]

\[= - 2a^{3} - 6\]

\[y_{2}(0) = - 2b^{3} - 6.\]

\[Точки\text{\ B\ }и\ D:\]

\[3xa^{2} - 2a^{3} - 6 = 0\]

\[3xa^{2} = 2a^{3} + 6\]

\[x_{a} = \frac{2}{3a} + \frac{2}{a^{2}} = \frac{2}{3} \bullet \left( a + \frac{3}{a^{2}} \right) =\]

\[= \frac{2}{3} \bullet \frac{a^{3} + 3}{a^{2}}\]

\[x_{b} = \frac{2}{3} \bullet \frac{b^{3} + 3}{b^{2}}.\]

\[Площади\ треугольников:\]

\[S_{\text{AOB}} = \frac{1}{2} \bullet \left( - 2a^{3} - 6 \right) \bullet \frac{2}{3} \bullet \frac{a^{3} + 3}{a^{2}} =\]

\[= \frac{a^{3} + 3}{3a^{2}} \bullet \left( - 2a^{3} - 6 \right) =\]

\[= \frac{- 2a^{6} - 6a^{3} - 6a^{3} - 18}{3a^{2}} =\]

\[= - \frac{2}{3} \bullet \frac{a^{6} + 6a^{3} + 9}{a^{2}} =\]

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

\[S_{\text{COD}} = - \frac{2}{3} \bullet \frac{\left( b^{3} + 3 \right)^{2}}{b^{2}}.\]

\[По\ условию:\]

\[\left\{ \begin{matrix} 3a^{2} = 3b^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ - \frac{2}{3} \bullet 4 \bullet \frac{\left( a^{3} + 3 \right)^{2}}{a^{2}} = - \frac{2}{3} \bullet \frac{\left( b^{3} + 3 \right)^{2}}{b^{2}} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a = - b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{{4\left( a^{3} + 3 \right)}^{2}}{a^{2}} = \frac{\left( b^{3} + 3 \right)^{2}}{b^{2}} \\ \end{matrix} \right.\ \]

\[\frac{{4\left( ( - b)^{3} + 3 \right)}^{2}}{( - b)^{2}} = \frac{\left( b^{3} + 3 \right)^{2}}{b^{2}}\]

\[\frac{{4\left( 3 - b^{3} \right)}^{2}}{b^{2}} = \frac{\left( b^{3} + 3 \right)^{2}}{b^{2}}\]

\[4\left( 3 - b^{3} \right)^{2} = \left( b^{3} + 3 \right)^{2}\]

\[36 - 24b^{3} + 4b^{6} = b^{6} + 6b^{3} + 9\]

\[3b^{6} - 30b^{3} + 27 = 0\]

\[b^{6} - 10b^{3} + 9 = 0\]

\[y = b^{3}:\]

\[y^{2} - 10y + 9 = 0\]

\[D = 100 - 36 = 64\]

\[y_{1} = \frac{10 - 8}{2} = 1\text{\ \ }\]

\[y_{2} = \frac{10 + 8}{2} = 9\]

\[b_{1} = \sqrt[3]{1} = 1\text{\ \ }b_{2} = \sqrt[3]{9}\]

\[a_{1} = - 1\ \ \ \ \ \ \ \ \text{\ \ }a_{2} = - \sqrt[3]{9}.\]

\[S_{\text{AOB}} = - \frac{2}{3} \bullet \frac{\left( ( - 1)^{3} + 3 \right)^{2}}{( - 1)^{2}} =\]

\[= - \frac{2}{3} \bullet 2^{2} = - \frac{8}{3} = \frac{8}{3}\]

\[S_{\text{AOB}} = - \frac{2}{3} \bullet \frac{\left( \left( - \sqrt[3]{9} \right)^{3} + 3 \right)^{2}}{\left( - \sqrt[3]{9} \right)^{2}} =\]

\[= - \frac{2}{3} \bullet \frac{( - 9 + 3)^{2}}{\sqrt[3]{9^{2}}} =\]

\[= - \frac{2}{3} \bullet \frac{36}{3\sqrt[3]{3}} = - \frac{8}{\sqrt[3]{3}} = \frac{8}{\sqrt[3]{3}}.\]

\[Ответ:\ \ \frac{8}{3}\ или\ \frac{8}{\sqrt[3]{3}}.\]