Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\inline \LARGE \dpi{110}\frac{\sqrt 2 - \sqrt 1}{\sqrt{2.1}}+\frac{\sqrt 3 - \sqrt 2}{\sqrt{3.2}}+\frac{\sqrt 4 - \sqrt 3}{\sqrt{4.3}}$ is gelijk aan	A. 1
B.
C.
D.
E. een irrationaal getal

$https://latex.codecogs.com/png.image?\inline \LARGE \dpi{110}\frac{\sqrt 2 - \sqrt 1}{\sqrt{2.1}}+\frac{\sqrt 3 - \sqrt 2}{\sqrt{3.2}}+\frac{\sqrt 4 - \sqrt 3}{\sqrt{4.3}}$ is equal to	A. 1
	B.
	C.
	D.
	E. an irrational number

1^ste manier :
$\frac{\sqrt 2-\sqrt 1}{\sqrt{2.1}}+\frac{\sqrt 3-\sqrt 2}{\sqrt{3.2}}+\frac{\sqrt 4-\sqrt 3}{\sqrt{4.3}}\\=\frac{\sqrt2-\sqrt1}{\sqrt2}+\frac{\sqrt3-\sqrt2}{\sqrt6}+\frac{\sqrt4-\sqrt3}{\sqrt{12}}\\=\frac{(\sqrt2-\sqrt1)\sqrt2}{2}+\frac{(\sqrt3-\sqrt2)\sqrt6}{6}+\frac{(\sqrt4-\sqrt3)\sqrt{12}}{12}\\=\frac{2-\sqrt2}{2}+\frac{\sqrt{18}-\sqrt{12}}{6}+\frac{\sqrt{48}-6}{12}\\=\frac{2-\sqrt2}{2}+\frac{3\sqrt2-2\sqrt3}{6}+\frac{4\sqrt3-6}{12}\\=1-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}-\frac{\sqrt3}{3}+\frac{\sqrt3}{3}-\frac12=\frac12$
2^de manier : "elegantste" manier
$\frac{\sqrt 2-\sqrt 1}{\sqrt{2.1}}+\frac{\sqrt 3-\sqrt 2}{\sqrt{3.2}}+\frac{\sqrt 4-\sqrt 3}{\sqrt{4.3}}\\=\frac{\sqrt2}{\sqrt2.\sqrt1}-\frac{\sqrt1}{\sqrt2.\sqrt1}+\frac{\sqrt3}{\sqrt3.\sqrt2}-\frac{\sqrt2}{\sqrt3.\sqrt2}+\frac{\sqrt4}{\sqrt4.\sqrt31}-\frac{\sqrt3}{\sqrt4-\sqrt3}\\=\frac1{\sqrt1}-\frac1{\sqrt2}+\frac1{\sqrt2}-\frac1{\sqrt3}+\frac1{\sqrt3}-\frac1{\sqrt4}\\=\frac1{\sqrt1}-\frac1{\sqrt4}=1-\frac12=\frac12=$