Упр.731 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1) (-3+2i)(1/3 — 1/2*i);
2) (-5 + корень 2i)(-6 — 3 корень 2i);
3) (1+i)(-1+2i) + 1-3i;
4) (3-2i)(4+i)+10i;
5) (3-i)(1+3i)/2-i;
6) 2-3i/(1-i)(3+i);
7) 3/2-3i + 3/2+3i;
8) 2-3i/2+i + 2+3i/2-i.

1. $$(-3+2i)\left(\frac13-\frac12 i\right)=-1+\frac32 i+\frac23 i-i^2$$
   $$=-1+\frac{9i+4i}{6}+1=\frac{13}{6}i.$$
2. $$(-5+\sqrt2\,i)(-6-3\sqrt2\,i)=30+15\sqrt2\,i+6\sqrt2\,i-6i^2$$
   $$=30+21\sqrt2\,i+6=36+21\sqrt2\,i.$$
3. $$(1+i)(-1+2i)+1-3i=-1+2i-i+2i^2+1-3i$$
   $$=-2-2i.$$
4. $$(3-2i)(4+i)+10i=12+3i-8i-2i^2+10i$$
   $$=12+5i+2=14+5i.$$
5. $$\frac{(3-i)(1+3i)}{2-i}=\frac{3+9i-i-3i^2}{2-i}=\frac{6+8i}{2-i}$$
   $$=\frac{(6+8i)(2+i)}{(2-i)(2+i)}=\frac{12+6i+16i+8i^2}{5}$$
   $$=\frac{4+22i}{5}=\frac45+\frac{22}{5}i.$$
6. $$\frac{2-3i}{(1-i)(3+i)}=\frac{2-3i}{3+i-3i-i^2}=\frac{2-3i}{4-2i}$$
   $$=\frac{(2-3i)(4+2i)}{(4-2i)(4+2i)}=\frac{8+4i-12i-6i^2}{16+4}$$
   $$=\frac{14-8i}{20}=\frac{7}{10}-\frac25 i.$$
7. $$\frac{3}{2-3i}+\frac{3}{2+3i}=\frac{3(2+3i)}{(2-3i)(2+3i)}+\frac{3(2-3i)}{(2+3i)(2-3i)}$$
   $$=\frac{6+9i+6-9i}{4-9i^2}=\frac{12}{13}.$$
8. $$\frac{2-3i}{2+i}+\frac{2+3i}{2-i}=\frac{(2-3i)(2-i)+(2+3i)(2+i)}{(2+i)(2-i)}$$
   $$=\frac{4-2i-6i+3i^2+4+2i+6i+3i^2}{4-i^2}$$
   $$=\frac{8-6}{5}=\frac25.$$

Ответ

1) $$\frac{13}{6}i$$; 2) $$36+21\sqrt2\,i$$; 3) $$-2-2i$$; 4) $$14+5i$$; 5) $$\frac45+\frac{22}{5}i$$; 6) $$\frac{7}{10}-\frac25 i$$; 7) $$\frac{12}{13}$$; 8) $$\frac25$$.