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Решите задачу:

$\displaystyle \frac{4i+5}{9-2i}-\frac{7}{4i+3}=\frac{(4i+5)(4i+3)}{(9-2i)(4i+3)}-\frac{7(9-2i)}{(4i+3)(9-2i)}=\\\\\\=\frac{-16+12i+20i+15-63+14i}{(4i+3)(9-2i)}=\frac{-64+46i}{36i+8+27-6i}=\frac{-64+46i}{35+30i}=\\\\\\\frac{2(-32+23i)}{5(7+6i)}=\frac{2(-32+23i)(7-6i)}{5(7+6i)(7-6i)}=\frac{2(-224+192i+161i+138)}{5(49-42i+42i+36)}\\\\\\=\frac{2(-86+353i)}{5*85}=\frac{-172+706i}{425}=-\frac{172}{425}+\frac{706}{425}i$