Question

Can anyone please answer this
$\frac{ {2}^{2001} + {2}^{1999} } { {2}^{2000} - {2}^{1998} }$
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Answer 1

Answer:

answer is 10/3

Step-by-step explanation:

hope it helps

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Answer 2

Step-by-step explanation:

$\\ \tt \implies\frac{ {2}^{2001} + {2}^{1999} } { {2}^{2000} - {2}^{1998} }$

Taking common,

$\[ \begin{array}{l} \tt =\dfrac{2^{1999}\left(2^{2}+2^{0}\right)}{2^{1998}\left(2^{2}-2^{0}\right)} \\ \\ \tt=\dfrac{2^{1999}(4+1)}{2^{1998}(4-1)} \\ \\ =\dfrac{2^{1999}(5)}{2^{1998}(3)} \\ \\ =\dfrac{5}{3} \cdot 2^{1999-1998} \quad\left[a^{b} \div a^{c}=a^{b-c}\right] \\ \\ \tt =\dfrac{5}{3} \cdot 2^{1} \\ \\ \tt =\dfrac{10}{3} \end{array} \]$