Đáp án:
Ta có :
2019 + $\dfrac{2018}{2}$ + .... + $\dfrac{2}{2018}$ + $\dfrac{1}{2019}$
= ( 1 + 1 + ... + 1) + $\dfrac{2018}{2}$ + .... + $\dfrac{2}{2018}$ + $\dfrac{1}{2019}$ ( 2019 số 1)
= 1 + ( 1 + $\dfrac{2018}{2}$)+ .... + ( 1 + $\dfrac{2}{2018}$ )+ ( 1 + $\dfrac{1}{2019}$)
= 1 + $\dfrac{2020}{2}$ + .... + $\dfrac{2020}{2018}$ + $\dfrac{2020}{2019}$
= $\dfrac{2020}{2020}$ +$\dfrac{2020}{2}$ + .... + $\dfrac{2020}{2018}$ + $\dfrac{2020}{2019}$
= 2020 . ( $\dfrac{1}{2}$ + .... + $\dfrac{1}{2019}$ + $\dfrac{1}{2020}$)
=> ( $\dfrac{1}{2}$ + .... + $\dfrac{1}{2019}$ + $\dfrac{1}{2020}$).x = 2019 + $\dfrac{2018}{2}$ + .... + $\dfrac{2}{2018}$ + $\dfrac{1}{2019}$
<=> ( $\dfrac{1}{2}$ + .... + $\dfrac{1}{2019}$ + $\dfrac{1}{2020}$) . x = 2020 . ( $\dfrac{1}{2}$ + .... + $\dfrac{1}{2019}$ + $\dfrac{1}{2020}$)
$<=> x = 2020$
Giải thích các bước giải:
