Question

алгебра номер 17.5(3) помогите пожалуйста ​

Answer 1

Ответ:

                    $\boxed{\ i^2=-1\ \ ,\ \ (a-b)(a+b)=a^2-b^2\ }$  

$1)\ \ \displaystyle \Big(\frac{3-i}{2+i}\Big)^2+(1-2i)^3=\Big(\frac{(3-i)(2-i)}{(2+i)(2-i)}\Big)^2+(1-3\cdot 2i+3\cdot 4i^2-8i^3)=\\\\\\=\Big(\frac{6-5i+i^2}{4-i^2}\Big)^2+1-6i-12-8\cdot \underbrace{i^2}_{-1}\cdot \, i=\\\\\\=\Big(\frac{5-5i}{4+1}\Big)^2-11-6i+8i=\frac{25-50i+25i^2}{5^2}-11+2i=\\\\\\=1-2i+\underbrace{i^2}_{-1}-11+2i=-11$

$3)\ \ \displaystyle (2+3i)^4-\frac{2-3i}{1+i}=(2+3i)^2\cdot (2+3i)^2-\frac{(2-3i)(1-i)}{(1+i)(1-i)}=\\\\\\=(4+12i+9i^2)(4+12i+9i^2)-\frac{2-2i-3i+3i^2}{1-i^2}=\\\\\\=(-5+12i)(-5+12i)-\frac{-1-5i}{1+1}=25-120i+144i^2+\frac{1+5i}{2}=\\\\\\=25-120i-144+0,5+2,5i=-118,5-117,5i$