Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\frac{x}{\sqrt x}+\frac{\sqrt x}{x}$ kan ook geschreven worden als	A. $\boldsymbol{2\,\sqrt x }$
	B. $\boldsymbol{2 }$
	C. $\boldsymbol{x\sqrt x}$
	D. $\large\boldsymbol{\frac {x^2\,+\,1} {\sqrt x} }$
	E. $\large\boldsymbol{\frac {x\,+\,1} {\sqrt x} }$

[ 3,4-7313 - op net sinds 29.11.13-(E)-3.11.2023 ]

Translation in E N G L I S H

$https://latex.codecogs.com/png.image?\dpi{110}\frac{x}{\sqrt x}+\frac{\sqrt x}{x}$ can also be written as	A. $\boldsymbol{2\,\sqrt x }$
	B. $\boldsymbol{2 }$
	C. $\boldsymbol{x\sqrt x}$
	D. $\large\boldsymbol{\frac {x^2\,+\,1} {\sqrt x} }$
	E. $\large\boldsymbol{\frac {x\,+\,1} {\sqrt x} }$

Oplossing - Solution

1^ste manier :
$\\\frac{x}{\sqrt x}+\frac{\sqrt x}{x}=\sqrt x + \frac{\sqrt x}{x}=\sqrt x\left ( 1+\frac 1x \right )=\sqrt x\cdot \frac{x+1}{x} =\sqrt x\cdot\frac{x+1}{\sqrt x.\sqrt x}=\frac{x+1}{\sqrt x}$
2^de manier :
$\\\frac{x}{\sqrt x}+\frac{\sqrt x}{x} =\frac{x^2+x}{x.\sqrt x}=\frac{x(x+1)}{x\sqrt x}=\frac{x+1}{\sqrt x}$
3^de manier :
$\\\frac{x}{\sqrt x}+\frac{\sqrt x}{x}=\frac{x}{\sqrt x}+\frac{\sqrt x}{\sqrt x.\sqrt x}=\frac{x}{\sqrt x}+\frac{1}{\sqrt x}=\frac{x+1}{\sqrt x}$

$https://latex.codecogs.com/png.image?\dpi{110}\frac{x}{\sqrt x}+\frac{\sqrt x}{x}$ kan ook geschreven worden als	A. \(\boldsymbol{2\,\sqrt x }\)
	B. \(\boldsymbol{2 }\)
	C. \(\boldsymbol{x\sqrt x}\)
	D. \(\large\boldsymbol{\frac {x^2\,+\,1} {\sqrt x} }\)
	E. \(\large\boldsymbol{\frac {x\,+\,1} {\sqrt x} }\)

$https://latex.codecogs.com/png.image?\dpi{110}\frac{x}{\sqrt x}+\frac{\sqrt x}{x}$ can also be written as	A. \(\boldsymbol{2\,\sqrt x }\)
	B. \(\boldsymbol{2 }\)
	C. \(\boldsymbol{x\sqrt x}\)
	D. \(\large\boldsymbol{\frac {x^2\,+\,1} {\sqrt x} }\)
	E. \(\large\boldsymbol{\frac {x\,+\,1} {\sqrt x} }\)