Giải thích các bước giải:
Ta có :
$\dfrac{1}{1.2}+\dfrac{1}{3.4}+..+\dfrac{1}{99.100}$
$=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+..+\dfrac{100-99}{99.100}$
$=\dfrac11-\dfrac12+\dfrac13-\dfrac14+..+\dfrac{1}{99}-\dfrac{1}{100}$
$=(\dfrac{1}{1}+\dfrac{1}{3}+..+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{100})$
$=(\dfrac{1}{1}+\dfrac{1}{3}+..+\dfrac{1}{99}+(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{100}))-2(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{100})$
$=(\dfrac{1}{1}+\dfrac12+\dfrac{1}{3}+..+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{1}+\dfrac{1}{2}+..+\dfrac{1}{50})$
$=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}$
$\to (\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}).2019x=\dfrac{2018}{51}+\dfrac{2018}{52}+\dfrac{2018}{53}+...+\dfrac{2018}{99}+\dfrac{2018}{100}$
$\to (\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}).2019x=2018.(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100})$
$\to x=\dfrac{2018}{2019}$
