ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 504

\[\left( \sqrt[3]{x} + \frac{1}{\sqrt{x}} \right)^{12}\]

\[1)\ a = C_{n}^{k} \bullet \left( \sqrt[3]{x} \right)^{n - k} \bullet \left( \frac{1}{\sqrt{x}} \right)^{k} =\]

\[= C_{12}^{k} \bullet \left( \sqrt[3]{x} \right)^{12 - k} \bullet \left( \sqrt{x} \right)^{- k} =\]

\[= C_{12}^{k} \bullet x^{\frac{1}{3}(12 - k)} \bullet x^{\frac{1}{2}( - k)} =\]

\[= C_{12}^{k} \bullet x^{4 - \frac{5}{6}k}.\]

\[2)\ 4 - \frac{5}{6}k = - 1\]

\[\frac{5}{6}k = 5\]

\[k = 6.\]

\[3)\ C_{12}^{6} = \frac{12!}{6!(12 - 6)!} =\]

\[= \frac{12 \bullet 11 \bullet 10 \bullet 9 \bullet 8 \bullet 7 \bullet 6!}{6 \bullet 5 \bullet 4 \bullet 3 \bullet 2 \bullet 6!} = 924.\]

\[Ответ:\ \ C_{12}^{6} \bullet \frac{1}{x} = \frac{924}{x}.\]