Rozwiązane

a)(-4i 2/3*(-2i2/7)-3i1/3*(0,4-3i1/5)=
b)(-3i3/4)*(-2i2/3-1i1/2:(-3²)=



Odpowiedź :

Problematyyyyyk
a)(-4i 2/3*(-2i2/7)-3i1/3*(0,4-3i1/5)= 6 i 2/15
a)=(-14/3)*(-16/7)-10/3*(4/10-16/5)=32/3-10/3*(4/10-32/10)=32/3-10/3*(-28/10)=32/3+28/3=60/3=20
b)=(-15/4)*(-8/3-3/2*1/9)=(-15/4)*(-8/3-1/6)=(-15/4)*(-16/6-1/6)=-15/4*(-17/6)=255/24=10i15/24

14/3 oznacza ułamek
Annaa300
a)(-4i 2/3*(-2i2/7)-3i1/3*(0,4-3i1/5)=zamieniam na niewlas ulamki
=(-14/3)*(-16/7)-10/3*(4/10-3i 2/10)=
=32/3-10/3*(4/10-32/10)=
=32/3-10/3*(-28/10)=
=32/3+28/3=
=60/3=
=20
b)(-3i3/4)*(-2i2/3-1i1/2:(-3²)=
=(-15/4)*(-8/3-3/2*1/9)=
=(-15/4)*(-8/3-1/6)=
=(-15/4)*(-16/6-1/6)=
=-15/4*(-17/6)=
=255/24=
=10i15/24
=10i 5/8

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