-
+1
Ovi korisnici su zahvalili autoru
Daniel za post:
lolica
Reputacija: 4.35%
od Daniel » Utorak, 06. Avgust 2013, 19:34
[inlmath]1.[/inlmath] zadatak:
Iskoristimo formulu za razliku kvadrata:
[dispmath]x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\quad\Rightarrow\quad x-y=\frac{x^3-y^3}{x^2+xy+y^2}[/dispmath]
pa zatim ovo primenimo na prvi faktor u zadatom izrazu:
[dispmath]\left(\sqrt[3]\frac{a}{b}-\sqrt[3]\frac{b}{a}\right)\left(\sqrt[3]\frac{a^2}{b^2}+\sqrt[3]\frac{b^2}{a^2}+1\right)=\frac{\left(\sqrt[3]\frac{a}{b}\right)^3-\left(\sqrt[3]\frac{b}{a}\right)^3}{\left(\sqrt[3]\frac{a}{b}\right)^2+\sqrt[3]{\frac{\cancel a}{\cancel b}\cdot\frac{\cancel b}{\cancel a}}+\left(\sqrt[3]\frac{b}{a}\right)^2}\left(\sqrt[3]\frac{a^2}{b^2}+\sqrt[3]\frac{b^2}{a^2}+1\right)=[/dispmath]
[dispmath]=\frac{\frac{a}{b}-\frac{b}{a}}{\cancel{\sqrt[3]\frac{a^2}{b^2}+\sqrt[3]\frac{b^2}{a^2}+1}}\cancel{\left(\sqrt[3]\frac{a^2}{b^2}+\sqrt[3]\frac{b^2}{a^2}+1\right)}=\frac{a}{b}-\frac{b}{a}=\frac{a^2-b^2}{ab}[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain