Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1094

Авторы:
Год:2020-2021-2022-2023
Тип:учебник
Серия:Алгебра и начала математического анализа, геометрия

Задание 1094

\[\boxed{\mathbf{1094}\mathbf{.}}\]

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

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

\[= \frac{12!}{(12 - 3)! \bullet 3!} \bullet \left( \sqrt{x} \right)^{9} \bullet x^{3} =\]

\[= \frac{12!}{9! \bullet 3!} \bullet x^{4}\sqrt{x} \bullet x^{3} =\]

\[= \frac{12 \bullet 11 \bullet 10 \bullet 9!}{9! \bullet 3 \bullet 2} \bullet x^{7} \bullet x^{\frac{1}{2}} =\]

\[= 4 \bullet 11 \bullet 5 \bullet x^{\frac{14}{2}} \bullet x^{\frac{1}{2}} = 220x^{\frac{15}{2}}\]

\[2)\ \left( x - \sqrt{x} \right)^{14}\]

\[C_{14}^{3} \bullet x^{14 - 3} \bullet \left( - \sqrt{x} \right)^{3} =\]

\[= \frac{14!}{(14 - 3)! \bullet 3!} \bullet x^{11} \bullet \left( - x^{\frac{1}{2}} \right)^{3} =\]

\[= - \frac{14!}{11! \bullet 3!} \bullet x^{\frac{22}{2}} \bullet x^{\frac{3}{2}} =\]

\[= - \frac{14 \bullet 13 \bullet 12 \bullet 11!}{11! \bullet 3 \bullet 2} \bullet x^{\frac{25}{2}} =\]

\[= - 14 \bullet 13 \bullet 2 \bullet x^{\frac{25}{2}} = - 364x^{\frac{25}{2}}\]

\[3)\ \left( x - \frac{1}{x} \right)^{13}\]

\[C_{13}^{3} \bullet x^{13 - 3} \bullet \left( - \frac{1}{x} \right)^{3} =\]

\[= \frac{13!}{(13 - 3)! \bullet 3!} \bullet x^{10} \bullet \left( - \frac{1}{x^{3}} \right) =\]

\[= - \frac{13!}{10! \bullet 3!} \bullet x^{7} =\]

\[= - \frac{13 \bullet 12 \bullet 11 \bullet 10!}{10! \bullet 3 \bullet 2} \bullet x^{7} =\]

\[= - 13 \bullet 2 \bullet 11 \bullet x^{7} = - 286x^{7}\]

\[4)\ \left( \frac{1}{x} + x \right)^{11}\]

\[C_{11}^{3} \bullet \left( \frac{1}{x} \right)^{11 - 3} \bullet x^{3} =\]

\[= \frac{11!}{(11 - 3)! \bullet 3!} \bullet \left( \frac{1}{x} \right)^{8} \bullet x^{3} =\]

\[= \frac{11!}{8! \bullet 3!} \bullet \frac{1}{x^{8}} \bullet x^{3} =\]

\[= \frac{11 \bullet 10 \bullet 9 \bullet 8!}{8! \bullet 3 \bullet 2} \bullet x^{- 5} =\]

\[= 11 \bullet 5 \bullet 3 \bullet x^{- 5} = 165x^{- 5}\]

\[5)\ \left( a^{0,1} + a^{0,2} \right)^{9}\]

\[C_{9}^{3} \bullet \left( a^{0,1} \right)^{9 - 3} \bullet \left( a^{0,2} \right)^{3} =\]

\[= \frac{9!}{(9 - 3)! \bullet 3!} \bullet \left( a^{0,1} \right)^{6} \bullet a^{0,6} =\]

\[= \frac{9!}{6! \bullet 3!} \bullet a^{0,6} \bullet a^{0,6} =\]

\[= \frac{9 \bullet 8 \bullet 7 \bullet 6!}{6! \bullet 3 \bullet 2} \bullet a^{1,2} =\]

\[= 3 \bullet 4 \bullet 7 \bullet a^{1,2} = 84a^{1,2}\]

\[6)\ \left( b^{0,3} + b^{0,4} \right)^{8}\]

\[C_{8}^{3} \bullet \left( b^{0,3} \right)^{8 - 3} \bullet \left( b^{0,4} \right)^{3} =\]

\[= \frac{8!}{(8 - 3)! \bullet 3!} \bullet \left( b^{0,3} \right)^{5} \bullet b^{1,2} =\]

\[= \frac{8!}{5! \bullet 3!} \bullet b^{1,5} \bullet b^{1,2} =\]

\[= \frac{8 \bullet 7 \bullet 6 \bullet 5!}{5! \bullet 3 \bullet 2} \bullet b^{2,7} =\]

\[= 4 \bullet 7 \bullet 2 \bullet b^{2,7} = 56b^{2,7}\]

Скачать ответ
Есть ошибка? Сообщи нам!

Решебники по другим предметам