Izraz sa korenima – prijemni GRF 2019.

PostPoslato: Ponedeljak, 08. Jul 2019, 19:22
od Stefan Boricic
Prijemni ispit GRF – 25. jun 2019.
1. zadatak


Vrednost izraza [inlmath]3\left(\sqrt5-\sqrt2\right)^{-1}+\left(\sqrt5-\sqrt2\right)[/inlmath] jednaka je:
[inlmath]A)\;\sqrt5+\sqrt2;\quad[/inlmath] [inlmath]\enclose{circle}{B)}\;2\sqrt5;\quad[/inlmath] [inlmath]C)\;2\sqrt2;\quad[/inlmath] [inlmath]D)\;5\sqrt2;\quad[/inlmath] [inlmath]E)\;\sqrt5-\sqrt2;\qquad[/inlmath] [inlmath]N)\;\text{Ne znam.}[/inlmath]

Rešavamo izraz:
[dispmath]3\left(\sqrt5-\sqrt2\right)^{-1}+\left(\sqrt5-\sqrt2\right)\\
3\cdot\frac{1}{\sqrt5-\sqrt2}+\sqrt5-\sqrt2\\
\frac{3}{\sqrt5-\sqrt2}+\sqrt5-\sqrt2[/dispmath] Racionališemo i dobijemo:
[dispmath]\frac{3\left(\sqrt5+\sqrt2\right)}{3}+\sqrt5-\sqrt2\\
\frac{\cancel{3}\left(\sqrt5+\sqrt2\right)}{\cancel3}+\sqrt5-\sqrt2\\
\sqrt5\cancel{+\sqrt2}+\sqrt5\cancel{-\sqrt2}\\
\enclose{box}{2\sqrt5}[/dispmath]