MATEMANIJA

15. Ako je [inlmath]i^2=-1[/inlmath] i [inlmath]\varepsilon[/inlmath] kompleksan broj koji zadovoljava uslov [inlmath]\varepsilon^2+\varepsilon+1=0[/inlmath], tada je rešenje jednačine [inlmath]\frac{x-1}{x+1}=\varepsilon\frac{1+i}{1-i}[/inlmath] po [inlmath]x[/inlmath], jednako:
[inlmath]\left(A\right)\;-2\varepsilon+1-2i\quad[/inlmath] [inlmath]\left(B\right)\;-2\varepsilon-1+2i\quad[/inlmath] [inlmath]\left(C\right)\;-2\varepsilon-1-2i\quad[/inlmath] [inlmath]\enclose{box}{\left(D\right)\;2\varepsilon+1-2i}\quad[/inlmath] [inlmath]\left(E\right)\;2\varepsilon-1-2i\quad[/inlmath] [inlmath]\left(N\right)\;\mbox{Ne znam}[/inlmath]

[dispmath]\frac{x-1}{x+1}=\varepsilon\frac{1+i}{1-i}[/dispmath]
prvo postavljamo uslov
[dispmath]x+1\ne 0\quad\Rightarrow\quad x\ne -1[/dispmath]
pa se oslobodimo razlomaka:
[dispmath]\left(x-1\right)\left(1-i\right)=\varepsilon\left(x+1\right)\left(1+i\right)[/dispmath][dispmath]x-1-ix+i=\varepsilon x+\varepsilon+i\varepsilon x+i\varepsilon[/dispmath]
Sabirke s [inlmath]x[/inlmath] grupišemo na jednu stranu, slobodne sabirke na drugu:
[dispmath]x-ix-\varepsilon x-i\varepsilon x=\varepsilon+i\varepsilon+1-i[/dispmath][dispmath]x\left(1-\varepsilon-i-i\varepsilon\right)=\varepsilon+1+i\varepsilon-i[/dispmath][dispmath]x=\frac{\left(\varepsilon+1\right)+i\left(\varepsilon-1\right)}{\left(1-\varepsilon\right)-i\left(1+\varepsilon\right)}[/dispmath]
To racionalizujemo:
[dispmath]x=\frac{\left(\varepsilon+1\right)+i\left(\varepsilon-1\right)}{\left(1-\varepsilon\right)-i\left(1+\varepsilon\right)}\cdot\frac{\left(1-\varepsilon\right)+i\left(1+\varepsilon\right)}{\left(1-\varepsilon\right)+i\left(1+\varepsilon\right)}[/dispmath]
[dispmath]x=\frac{\left(\varepsilon+1\right)\left(1-\varepsilon\right)-\left(\varepsilon-1\right)\left(1+\varepsilon\right)+i\left[\left(\varepsilon-1\right)\left(1-\varepsilon\right)+\left(\varepsilon+1\right)\left(1+\varepsilon\right)\right]}{\left(1-\varepsilon\right)^2+\left(1+\varepsilon\right)^2}[/dispmath]
[dispmath]x=\frac{-\left(\varepsilon+1\right)\left(\varepsilon-1\right)+\left(1-\varepsilon\right)\left(1+\varepsilon\right)+i\left[-\left(\varepsilon-1\right)\left(\varepsilon-1\right)+\left(\varepsilon+1\right)\left(\varepsilon+1\right)\right]}{1-2\varepsilon+\varepsilon^2+1+2\varepsilon+\varepsilon^2}[/dispmath]
[dispmath]x=\frac{-\left(\varepsilon^2-1\right)+\left(1-\varepsilon^2\right)+i\left[-\left(\varepsilon^2-2\varepsilon+1\right)+\left(\varepsilon^2+2\varepsilon+1\right)\right]}{2+2\varepsilon^2}[/dispmath]
[dispmath]x=\frac{2-2\varepsilon^2+i4\varepsilon}{2+2\varepsilon^2}[/dispmath]
skratimo sa [inlmath]2[/inlmath],
[dispmath]x=\frac{1-\varepsilon^2+i2\varepsilon}{1+\varepsilon^2}[/dispmath]
Iz [inlmath]\varepsilon^2+\varepsilon+1=0[/inlmath] sledi [inlmath]\varepsilon^2=-\varepsilon-1[/inlmath], pa to uvrstimo u izraz:
[dispmath]x=\frac{1-\left(-\varepsilon-1\right)+i2\varepsilon}{1+\left(-\varepsilon-1\right)}[/dispmath][dispmath]x=\frac{2+\varepsilon+i2\varepsilon}{-\varepsilon}[/dispmath][dispmath]x=\frac{2}{-\varepsilon}-1-i2[/dispmath][dispmath]\varepsilon^2+\varepsilon+1=0\quad\Rightarrow\quad\varepsilon^2+\varepsilon=-1\quad\Rightarrow\quad\varepsilon\left(\varepsilon+1\right)=-1\quad\Rightarrow\quad -\frac{1}{\varepsilon}=\varepsilon+1[/dispmath][dispmath]x=2\left(\varepsilon+1\right)-1-i2[/dispmath][dispmath]\underline{x=2\varepsilon+1-i2}[/dispmath]

Jel mozes i ovaj?
Oko polulopte poluprecnika [inlmath]r[/inlmath] opisana je prava kupa minimalne zapremine cija je osnova u ravni osnove polulopte. Zapremina te kupe je?

Kompleksni brojevi baš i nisu pravo mesto za polulopte i opisane kupe.

Odgovorio sam ti u „Geometriji“.

Joj izvini nisam ni razmisljao gde postavaljam, isao sam preko linka

Malo brzi nacin resavanja moze ici ako primetimo da je u jednacini:
[dispmath]\frac{x-1}{x+1}=\varepsilon\frac{1+i}{1-i}[/dispmath]
deo [inlmath]\displaystyle\frac{1+i}{1-i}[/inlmath] jednak [inlmath]i[/inlmath], pa se to odmah svede na:
[dispmath]\frac{x-1}{x+1}=i\varepsilon\\
\vdots[/dispmath]