Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\frac{\sqrt{21}+\sqrt6}{\sqrt3}$ is gelijk aan	A. +
B. +
C. 3
D. 9
E. +

$\frac{\sqrt{21}+\sqrt6}{\sqrt3}$ is equal to	A. +
B. +
C. 3
D. 9
E. +

1^ste manier :
$\frac{\sqrt{21}+\sqrt6}{\sqrt3}=\frac{\sqrt{21}}{\sqrt3}+\frac{\sqrt6}{\sqrt3}=\sqrt7+\sqrt2$
2^de manier :
$\frac{\sqrt{21}+\sqrt6}{\sqrt3}=\frac{(\sqrt{21}+\sqrt6)\sqrt3}{\sqrt3.\sqrt3}=\frac{\sqrt{21}.\sqrt3+\sqrt6.\sqrt3}{3}=\frac{\sqrt7.\sqrt3.\sqrt3+\sqrt2.\sqrt3.\sqrt3}{3}\\=\frac{3.\sqrt7+3\sqrt2}{3}=\sqrt7+\sqrt2$
3^de manier :
$\frac{\sqrt{21}+\sqrt6}{\sqrt3}=\frac{\sqrt{7.3}+\sqrt{2.3}}{\sqrt3}=\frac{(\sqrt7+\sqrt2)\sqrt3}{\sqrt3}=\sqrt7+\sqrt2$
4^de manier :
$\\\frac{\sqrt{21}+\sqrt6}{\sqrt3}=\frac{\sqrt{21}}{\sqrt3}+\frac{\sqrt6}{\sqrt3}=\frac{\sqrt{21}\sqrt3}{\sqrt3.\sqrt3}+\frac{\sqrt6.\sqrt3}{\sqrt3.\sqrt3}=\frac{\sqrt{63}}{3}+\frac{\sqrt{18}}{3}\\=\frac{3\sqrt7}{3}+\frac{3\sqrt2}{3}=\sqrt7+\sqrt2$