784.

Algebarski oblik

TEKST ZADATKA

Izračunati:

(4i3−1)12 \bigg(\frac4{i\sqrt3-1}\bigg)^{12}
REŠENJE ZADATKA

Da bi se uklonio imaginarni broj iz imenioca, pomnožiti brojilac i imenilac konjugovanim brojem od i3−1,i\sqrt3-1, što je −i3−1.-i\sqrt3-1.

(4i3−1⋅−i3−1−i3−1)12(4(−3i−1)1−3i2)12 \bigg(\frac4{i\sqrt3-1}\cdot\frac{-i\sqrt3-1}{-i\sqrt3-1}\bigg)^{12} \\ \bigg(\frac{4(-\sqrt3i-1)}{1-3i^2}\bigg)^{12}

Pošto je i2=−1i^2=-1 dobija se:

(4(−3i−1)1+3)12(4(−3i−1)4)12(−3i−1)12 \bigg(\frac{4(-\sqrt3i-1)}{1+3}\bigg)^{12}\\ \bigg(\frac{4(-\sqrt3i-1)}{4}\bigg)^{12}\\ (-\sqrt3i-1)^{12}

Primeniti formulu za stepen stepena: (am)n=am⋅n(a^m)^n=a^{m\cdot n}

((−3i−1)2)6(3i2+23i+1)6((-\sqrt3i-1)^2)^6 \\ (3i^2+2\sqrt3i+1)^6

Pošto je i2=−1i^2=-1 dobija se:

(−3+23i+1)6(23i−2)6(-3+2\sqrt3i+1)^6 \\ (2\sqrt3i-2)^6

Primeniti formulu za stepen stepena: (am)n=am⋅n(a^m)^n=a^{m\cdot n}

((23i−2)2)3(12i2−83i+4)3((2\sqrt3i-2)^2)^3 \\ (12i^2-8\sqrt3i+4)^3

Pošto je i2=−1i^2=-1 dobija se:

(−12−83i+4)3(−83i−8)3−512⋅33i3−3⋅192i2⋅8−3⋅83i⋅64−512−15363i3−4608i2−15363i−512(-12-8\sqrt3i+4)^3 \\ (-8\sqrt3i-8)^3 \\ -512\cdot3\sqrt3i^3-3\cdot192i^2\cdot8-3\cdot8\sqrt3i\cdot64-512 \\ -1536\sqrt3i^3-4608i^2-1536\sqrt3i-512

Pošto je i2=−1i^2=-1 i i3=−ii^3=-idobija se:

15363i+4608−15363i−51240961536\sqrt3i+4608-1536\sqrt3i-512\\ 4096

Balkan Tutor Sva Prava Zadržana © 2025

Politika privatnosti