5) 1) (2−8i)(1+3i)+13i; 2) 1−3i(11i²+14i³)−7i⁴; 3) i⁴+2i+(7+i)/(5+6i); 4) 200(i+9)(i⁴−5); 5) 3i+(1−7i)/(4+i)

1. \((2-8i)(1+3i)+13i=2+6i-8i-24i^2+13i=2-2i+24+13i=26+11i.\)

2. Используем \(i^2=-1\), \(i^3=-i\), \(i^4=1\):

   \(1-3i(11i^2+14i^3)-7i^4=1-3i(-11-14i)-7=1-(42-33i)-7=-48+33i.\)

3. \(i^4+2i+\displaystyle\frac{7+i}{5+6i}=1+2i+\displaystyle\frac{(7+i)(5-6i)}{61}.\)

   \((7+i)(5-6i)=35-42i+5i-6i^2=41-37i.\)

   Следовательно, \(1+2i+\displaystyle\frac{41-37i}{61}=\displaystyle\frac{102+85i}{61}.\)

4. \(200(i+9)(i^4-5)=200(i+9)(1-5)=-800(i+9)=-7200-800i.\)

5. \(3i+\displaystyle\frac{1-7i}{4+i}=3i+\displaystyle\frac{(1-7i)(4-i)}{17}.\)

   \((1-7i)(4-i)=4-i-28i+7i^2=-3-29i.\)

   Итак, \(3i+\displaystyle\frac{-3-29i}{17}=\displaystyle\frac{-3+22i}{17}.\)

Ответ: 1) \(26+11i\); 2) \(-48+33i\); 3) \(\frac{102+85i}{61}\); 4) \(-7200-800i\); 5) \(\frac{-3+22i}{17}\).