(2+i/2-3i)+(1-i/5+2i)

Решите задачу:

$\frac{2+i}{2-3i} + \frac{1-i}{5+2i}= \frac{(2+i)(2+3i)}{(2-3i)(2+3i)} + \frac{(1-i)(5-2i)}{(5+2i)(5-2i)}= \frac{(2+i)(2+3i)}{(2)^2-(3i)^2} + \frac{(1-i)(5-2i)}{5^2-(2i)^2}= \\ \\ =\frac{4+2i+6i-3i^2}{4+9} + \frac{5-5i-2i+2i^2}{25+4}=\frac{7+8i}{13} + \frac{3-7i}{29}= \frac{29(7+8i)+13(3-7i)}{13\cdot 29} = \\ \\$

$\frac{203+232i+39-91i}{13\cdot 29} = \frac{242+141i}{377}$