Tìm x, biết:
($\frac{1}{2}$ + $\frac{1}{3}$ + …….. + $\frac{1}{2020}$).x = $\frac{2019}{1}$ + $\frac{2018}{2}$ + ……. +$\frac{2}{2018}$ + $\frac{1}{2019}$
Tìm x, biết:
($\frac{1}{2}$ + $\frac{1}{3}$ + …….. + $\frac{1}{2020}$).x = $\frac{2019}{1}$ + $\frac{2018}{2}$ + ……. +$\frac{2}{2018}$ + $\frac{1}{2019}$
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 . ( $\dfrac{1}{2}$ + …. + $\dfrac{1}{2019}$ + $\dfrac{1}{2020}$)
$<=> x = 2020$
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