Question

(2^1999-2^1998-2^1997) : (4)^998

Answer 1

$( {2}^{1999} - {2}^{1998} - {2}^{1997} ) \div {4}^{998}$

$= [ {2}^{1997}( {2}^{2} - 2 - 1) ] \div { ({2}^{2} )}^{998}$

$= [ {2}^{1997} (4 - 2 - 1)] \div {2}^{2\times998}$

$= ( {2}^{1997} \times 1) \div {2}^{1996}$

$= {2}^{1997} \div {2}^{1996}$

$= {2}^{1997 - 1996}$

$= {2}^{1}$

$= 2$

Answer 2

$2^{1999}-2^{1998}-2^{1997}:4^{998} =\\ \\ 2^{1997}(2^{2} -2^{1} -1):(2^{2})^{998}=\\ \\ 2^{1997}(4-2-1):2^{1996} =\\ \\ 2^{1997}:2^{1996}=2^{1997-1996} \\ \\ =2^{1}\\ \\ =2$