Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\inline \LARGE \dpi{110}\frac{a^2+b^2}{a^2-b^2}-\frac{b}{a-b}=$	A. $\large\boldsymbol{\frac {a} {a\,+\,b} }$
	B. $\large\boldsymbol{\frac {a^2} {a^2\,-\,b^2} }$
	C. $\large\boldsymbol{\frac {a} {a\,-\,b} }$
	D. $\large\boldsymbol{-\:\frac {a} {a\,-\,b} }$
	E. $\small\boldsymbol{\text{geen van de vorige} }$

[ 3-0916 - op net sinds 11.1.15-(E)-3.11.2024 ]

Translation in E N G L I S H

See above E. None of the above

Oplossing - Solution

$\\ \frac{a^2+b^2}{a^2-b^2}-\frac{b}{a-b} =\frac{a^2+b^2}{(a-b)(a+b)} - \frac{b.(a+b)}{(a-b)(a+b)} =\frac{a^2+b^2-ab-b^2}{(a-b)(a+b)}\\ {\color{Green} .}\qquad =\frac{a^2-ab}{(a-b)(a+b)} =\frac{a.(a-b)}{(a-b)(a+b)} =\frac{a}{a+b}$
$\\ \frac{a^2+b^2}{a^2-b^2}-\frac{b}{a-b} =\frac{a^2+b^2}{(a-b)(a+b)} - \frac{b.(a+b)}{(a-b)(a+b)} =\frac{a^2+b^2-ab-b^2}{(a-b)(a+b)}\\ {\color{Green} .}\qquad =\frac{a^2-ab}{(a-b)(a+b)} =\frac{a.(a-b)}{(a-b)(a+b)} =\frac{a}{a+b}$

$https://latex.codecogs.com/png.image?\inline \LARGE \dpi{110}\frac{a^2+b^2}{a^2-b^2}-\frac{b}{a-b}=$	A. \(\large\boldsymbol{\frac {a} {a\,+\,b} }\)
	B. \(\large\boldsymbol{\frac {a^2} {a^2\,-\,b^2} }\)
	C. \(\large\boldsymbol{\frac {a} {a\,-\,b} }\)
	D. \(\large\boldsymbol{-\:\frac {a} {a\,-\,b} }\)
	E. \(\small\boldsymbol{\text{geen van de vorige} }\)