Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 660

\[1)\ \frac{10 + i^{3}}{- 5 + 2i} = \frac{10 - i}{2i - 5} =\]

\[= \frac{(10 - i)(2i + 5)}{(2i - 5)(2i + 5)} =\]

\[= \frac{20i + 50 - 2i^{2} - 5i}{4i^{2} - 25} =\]

\[= \frac{15i + 50 + 2}{- 4 - 25} = \frac{- 52 - 15i}{29} =\]

\[= - \frac{52}{29} - \frac{15}{29}i;\]

\[2)\ \frac{4 + 3i}{3 - 4i} - \frac{5 - 4i}{4 + 5i} =\]

\[= \frac{(4 + 3i)(3 + 4i)}{(3 - 4i)(3 + 4i)} - \frac{(5 - 4i)(4 - 5i)}{(4 + 5i)(4 - 5i)} =\]

\[= \frac{12 + 16i + 9i + 12i^{2}}{9 - 16i^{2}} - \frac{20 - 25i - 16i + 20i^{2}}{16 - 25i^{2}} =\]

\[= \frac{12 + 25i - 12}{9 + 16} - \frac{20 - 41i - 20}{16 + 25} =\]

\[= \frac{25i}{25} - \frac{- 41i}{41} = i + i = 2i;\]

\[3)\ i + \frac{1 + 6i}{1 - 7i} = i + \frac{(1 + 6i)(1 + 7i)}{(1 - 7i)(1 + 7i)} =\]

\[= i + \frac{1 + 6i + 7i + 42i^{2}}{1 - 49i^{2}} =\]

\[= i + \frac{1 + 13i - 42}{1 + 49} =\]

\[= i + \frac{- 41 + 13i}{50} =\]

\[= - \frac{41}{50} + \frac{13}{50}i + i = - \frac{41}{50} + \frac{63}{50}i;\]

\[4)\ \frac{5 + i}{(1 + i)(2 - 3i)} =\]

\[= \frac{5 + i}{2 - 3i + 2i - 3i^{2}} =\]

\[= \frac{5 + i}{2 - i + 3} = \frac{5 + i}{5 - i} =\]

\[= \frac{(5 + i)^{2}}{(5 - i)(5 + i)} = \frac{25 + 10i + i^{2}}{25 - i^{2}} =\]

\[= \frac{25 + 10i - 1}{25 + 1} = \frac{24 + 10i}{26} =\]

\[= \frac{12}{13} + \frac{5}{13}i.\]