Đáp án:
Giải thích các bước giải:
Bài 1:
$= \dfrac{(2002+ 1)× 14+ 1998+ 2001× 2002}{2002. (1+ 503+ 504)}$
$= \dfrac{2002× 14+ 14+ 1998+ 2001× 2002}{2002× 1008}$
$= \dfrac{2002× 14+ 2002× 2001+ 2002}{2002× 1008}$
$= \dfrac{2002× (14+ 2002+ 1}{2002× 2008}$
$= \dfrac{2002× 2016}{2002× 2008}$
$= \dfrac{2016}{1008}= \dfrac{2}{1}= 2$
Bài 2:
$= 1- \dfrac{1}{2}+ \dfrac{1}{2}- \dfrac{1}{4}+ \dfrac{1}{4}- \dfrac{1}{8}+ \dfrac{1}{8}- \dfrac{1}{16}+ \dfrac{1}{16}- \dfrac{1}{32}+ \dfrac{1}{32}- \dfrac{1}{64}+ \dfrac{1}{64}- \dfrac{1}{128}+ \dfrac{1}{128}$
$= 1- \dfrac{1}{128}$
$= \dfrac{128}{128}- \dfrac{1}{128}$
$= \dfrac{127}{128}$
