Kann jemand bitte folgende Aufgabe lösen und Schritt für Schritt den Lösungsweg anzeigen. Vielen Dank :) !! (2^199 - 2^198) / (2^199 + 2^198)

$${\frac{\left({{\mathtt{2}}}^{{\mathtt{199}}}{\mathtt{\,-\,}}{{\mathtt{2}}}^{{\mathtt{198}}}\right)}{\left({{\mathtt{2}}}^{{\mathtt{199}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{2}}}^{{\mathtt{198}}}\right)}} = \left({\frac{{{\mathtt{2}}}^{{\mathtt{198}}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({{\mathtt{2}}}^{{\mathtt{198}}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)\right)}}\right)$$  =  $${\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

$${{\mathtt{2}}}^{{\mathtt{199}}} = {\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{198}}}$$      Man kann   $${{\mathtt{2}}}^{{\mathtt{198}}}$$   ausklammern und dann kürzen !

Ergebnis deiner Potenzaufgabe  =  1/3 =  $${\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$  !

Zur Erläuterung:

$${\frac{\left({{\mathtt{a}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right)}{\left({{\mathtt{a}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right)}} = {\frac{{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)\right)}}$$  =  $${\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Gruß radix !