1,tính nhanh A=2/1×3+2/3×5+…+2/2015×2017+2/2017×2019 06/09/2021 Bởi Valerie 1,tính nhanh A=2/1×3+2/3×5+…+2/2015×2017+2/2017×2019
A = $\dfrac{2}{1 × 3}$ + $\dfrac{2}{3 × 5}$ + … + $\dfrac{2}{2015 × 2017}$ + $\dfrac{2}{2017 × 2019}$ = $\dfrac{1}{1}$ – $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{5}$ +… + $\dfrac{1}{2015}$ – $\dfrac{1}{2017}$ + $\dfrac{1}{2017}$ – $\dfrac{1}{2019}$ = $\dfrac{1}{1}$ – $\dfrac{1}{2019}$ = $\dfrac{2019 – 1}{2019}$ = $\dfrac{2018}{2019}$ Bình luận
Ta có: A = $\frac{2}{1.3}$ + $\frac{2}{3.5}$ +…+$\frac{2}{2017.2019}$ ⇔A=$\frac{3-1}{1.3}$ + $\frac{5-3}{3.5}$ +…+$\frac{2019-2017}{2017.2019}$ ⇔A=1 – $\frac{1}{3}$ +$\frac{1}{3}$ -$\frac{1}{5}$ +….+$\frac{1}{2017}$ -$\frac{1}{2019}$ ⇔A= 1 – $\frac{1}{2019}$ ⇔A=$\frac{2018}{2019}$ Bình luận
A = $\dfrac{2}{1 × 3}$ + $\dfrac{2}{3 × 5}$ + … + $\dfrac{2}{2015 × 2017}$ + $\dfrac{2}{2017 × 2019}$
= $\dfrac{1}{1}$ – $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{5}$ +… + $\dfrac{1}{2015}$ – $\dfrac{1}{2017}$ + $\dfrac{1}{2017}$ – $\dfrac{1}{2019}$
= $\dfrac{1}{1}$ – $\dfrac{1}{2019}$
= $\dfrac{2019 – 1}{2019}$
= $\dfrac{2018}{2019}$
Ta có:
A = $\frac{2}{1.3}$ + $\frac{2}{3.5}$ +…+$\frac{2}{2017.2019}$
⇔A=$\frac{3-1}{1.3}$ + $\frac{5-3}{3.5}$ +…+$\frac{2019-2017}{2017.2019}$
⇔A=1 – $\frac{1}{3}$ +$\frac{1}{3}$ -$\frac{1}{5}$ +….+$\frac{1}{2017}$ -$\frac{1}{2019}$
⇔A= 1 – $\frac{1}{2019}$
⇔A=$\frac{2018}{2019}$