ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1096

\[\boxed{\mathbf{1096}\mathbf{.}}\]

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

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

\[= C_{12}^{n} \bullet x^{4 - \frac{n}{3}} \bullet x^{- \frac{n}{2}} = C_{12}^{n} \bullet x^{4 - \frac{n}{3} - \frac{n}{2}}\]

\[Номер\ члена,\ содержащего\ x^{- 1}:\]

\[x^{4 - \frac{n}{3} - \frac{n}{2}} = x^{- 1}\]

\[4 - \frac{n}{3} - \frac{n}{2} = - 1\]

\[\frac{2n + 3n}{6} = 5\]

\[5n = 30\]

\[n = 6.\]

\[Биноминальный\ \]

\[коэффициент:\]

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

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

\[= 11 \bullet 2 \bullet 3 \bullet 2 \bullet 7 = 924.\]

\[Ответ:\ \ 924x^{- 1}.\]

\[2)\ \left( \sqrt{x} + \frac{1}{\sqrt[3]{x}} \right)^{16}\]

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

\[= C_{16}^{n} \bullet x^{8 - \frac{n}{2}} \bullet x^{- \frac{n}{3}} = C_{12}^{n} \bullet x^{8 - \frac{n}{3} - \frac{n}{2}}\]

\[Номер\ члена,\ содержащего\ x^{3}:\]

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

\[8 - \frac{n}{3} - \frac{n}{2} = 3\]

\[\frac{2n + 3n}{6} = 5\]

\[5n = 30\ \]

\[n = 6.\]

\[Биноминальный\ \]

\[коэффициент:\]

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

\[= \frac{16 \bullet 15 \bullet 14 \bullet 13 \bullet 12 \bullet 11 \bullet 10!}{10! \bullet 6 \bullet 5 \bullet 4 \bullet 3 \bullet 2} =\]

\[= 4 \bullet 14 \bullet 13 \bullet 11 = 8008.\]

\[Ответ:\ \ 8008x^{3}.\]