Упр.660 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

2) 4+3i/3-4i — 5-4i4+5i;
3) i + 1+6i/1-7i;
4) 5+i/(1+i)(2-3i).

1. $$\frac{10+i^3}{-5+2i}=\frac{10-i}{-5+2i}=\frac{(10-i)(-5-2i)}{(-5+2i)(-5-2i)}$$
   $$=\frac{-50-20i+5i+2i^2}{25-4i^2}=\frac{-52-15i}{29}=-\frac{52}{29}-\frac{15}{29}i.$$
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. $$i+\frac{1+6i}{1-7i}=i+\frac{(1+6i)(1+7i)}{(1-7i)(1+7i)}$$
   $$=i+\frac{1+13i+42i^2}{1-49i^2}=i+\frac{-41+13i}{50}$$
   $$=-\frac{41}{50}+\frac{13}{50}i+i=-\frac{41}{50}+\frac{63}{50}i.$$
4. $$\frac{5+i}{(1+i)(2-3i)}=\frac{5+i}{2+2i-3i-3i^2}=\frac{5+i}{5-i}$$
   $$=\frac{(5+i)(5+i)}{(5-i)(5+i)}=\frac{25+10i+i^2}{25-i^2}=\frac{24+10i}{26}$$
   $$=\frac{12}{13}+\frac{5}{13}i.$$

Ответ

1) $$-\frac{52}{29}-\frac{15}{29}i$$;
2) $$2i$$;
3) $$-\frac{41}{50}+\frac{63}{50}i$$;
4) $$\frac{12}{13}+\frac{5}{13}i$$.