Упр.601 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. Вычислить:
   1) $$\frac{(4-3i)(2-i)}{1+i}$$;
   2) $$\frac{(1+i)(2+i)}{3-i}$$;
   3) $$\frac{1-i}{(2+i)(3-4i)}$$;
   4) $$\frac{2-i}{(3-i)(1+3i)}$$;
   5) $$\frac{5+2i}{2-5i}+\frac{3-4i}{4+3i}$$;
   6) $$\frac{2-3i}{1+4i}-\frac{2+3i}{1-4i}$$.

1. $$\frac{(4-3i)(2-i)}{1+i}=\frac{8-4i-6i+3i^2}{1+i}=\frac{5-10i}{1+i}$$
   $$=\frac{(5-10i)(1-i)}{(1+i)(1-i)}=\frac{5-5i-10i+10i^2}{2}=\frac{-5-15i}{2}$$

2. $$\frac{(1+i)(2+i)}{3-i}=\frac{2+3i+i^2}{3-i}=\frac{1+3i}{3-i}$$
   $$=\frac{(1+3i)(3+i)}{(3-i)(3+i)}=\frac{3+i+9i+3i^2}{10}=\frac{10i}{10}=i$$

3. $$\frac{1-i}{(2+i)(3-4i)}=\frac{1-i}{6-8i+3i-4i^2}=\frac{1-i}{10-5i}$$
   $$=\frac{(1-i)(10+5i)}{(10-5i)(10+5i)}=\frac{10+5i-10i-5i^2}{100+25}=\frac{15-5i}{125}$$
   $$=\frac{3}{25}-\frac{1}{25}i$$

4. $$\frac{2-i}{(3-i)(1+3i)}=\frac{2-i}{3+9i-i-3i^2}=\frac{2-i}{6+8i}$$
   $$=\frac{(2-i)(6-8i)}{(6+8i)(6-8i)}=\frac{12-16i-6i+8i^2}{36+64}=\frac{4-22i}{100}$$
   $$=\frac{1}{25}-\frac{11}{50}i$$

5. $$\frac{5+2i}{2-5i}+\frac{3-4i}{4+3i}=\frac{(5+2i)(2+5i)}{(2-5i)(2+5i)}+\frac{(3-4i)(4-3i)}{(4+3i)(4-3i)}$$
   $$=\frac{10+25i+4i+10i^2}{4+25}+\frac{12-9i-16i+12i^2}{16+9}$$
   $$=\frac{29i}{29}+\frac{-25i}{25}=i-i=0$$

6. $$\frac{2-3i}{1+4i}-\frac{2+3i}{1-4i}=\frac{(2-3i)(1-4i)-(2+3i)(1+4i)}{(1+4i)(1-4i)}$$
   $$=\frac{2-8i-3i+12i^2-2-8i-3i-12i^2}{1-16i^2}=\frac{-22i}{17}$$

Ответ

1. $$-\frac{5}{2}-\frac{15}{2}i$$
2. $$i$$
3. $$\frac{3}{25}-\frac{1}{25}i$$
4. $$\frac{1}{25}-\frac{11}{50}i$$
5. $$0$$
6. $$-\frac{22}{17}i$$