ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 462

\[\left( \sqrt{x} + \frac{1}{\sqrt[3]{x}} \right)^{16}.\]

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

\[= C_{16}^{k} \bullet \left( \sqrt{x} \right)^{16 - k} \bullet \left( \sqrt[3]{x} \right)^{- k} =\]

\[= C_{16}^{k} \bullet x^{\frac{1}{2}(16 - k)} \bullet x^{\frac{1}{3}( - k)} =\]

\[= C_{16}^{k} \bullet x^{8 - \frac{k}{2}} \bullet x^{- \frac{k}{3}} = C_{16}^{k} \bullet x^{8 - \frac{5k}{6}}.\]

\[2)\ 8 - \frac{5k}{6} = 3\]

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

\[k = 6.\]

\[3)\ C_{16}^{6} = \frac{16!}{6!(10 - 6)!} =\]

\[= \frac{16 \bullet 15 \bullet 14 \bullet 13 \bullet 12 \bullet 10!}{6 \bullet 5 \bullet 4 \bullet 3 \bullet 2 \bullet 10!} =\]

\[= 8008.\]

\[Ответ:\ \ C_{16}^{6} \bullet x^{3} = 8008x^{3}.\]