$\frac{1}{1.3}$ + $\frac{1}{3.5}$+ $\frac{1}{7.9}$ +…+ $\frac{1}{2019.2021}$ Giúp với ạ! Giải thích rõ ràng 25/09/2021 Bởi Quinn $\frac{1}{1.3}$ + $\frac{1}{3.5}$+ $\frac{1}{7.9}$ +…+ $\frac{1}{2019.2021}$ Giúp với ạ! Giải thích rõ ràng
Đáp án: đề thiếu , em xem lại nha, anh sửa lại đề đó $\dfrac{1}{1.3}$ + $\dfrac{1}{3.5}$ + $\dfrac{1}{5.7}$ + $\dfrac{1}{7.9}$ + … + $\dfrac{1}{2019.2021}$ = $\dfrac{1}{2}$ . [1- $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{5}$ + $\dfrac{1}{5}$ – $\dfrac{1}{7}$ + $\dfrac{1}{7}$- $\dfrac{1}{9}$ + … + (-$\dfrac{1}{2019}$) + $\dfrac{1}{2019}$ – $\dfrac{1}{2021}$ ] = $\dfrac{1}{2}$ . [ 1 – $\dfrac{1}{2021}$] = $\dfrac{1}{2}$ . [ $\dfrac{2021}{2021}$ -$\dfrac{1}{2021}$ ] = $\dfrac{1}{2}$ . $\dfrac{2020}{2021}$ =$\dfrac{1010}{2021}$ Bình luận
Đề thiếu nha bạn. $\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + …. + \dfrac{1}{2019.2021}$ $= \dfrac{1}{2} . (\dfrac{2}{1.3} + \dfrac{2}{3.5} + \dfrac{2}{5.7} + …. + \dfrac{2}{2019.2021})$ $= \dfrac{1}{2} . (1 – \dfrac{1}{2021})$ $= \dfrac{1}{2} . \dfrac{2020}{2021}$ $= \dfrac{1010}{2021}$ Bình luận
Đáp án: đề thiếu , em xem lại nha, anh sửa lại đề đó
$\dfrac{1}{1.3}$ + $\dfrac{1}{3.5}$ + $\dfrac{1}{5.7}$ + $\dfrac{1}{7.9}$ + … + $\dfrac{1}{2019.2021}$
= $\dfrac{1}{2}$ . [1- $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{5}$ + $\dfrac{1}{5}$ – $\dfrac{1}{7}$ + $\dfrac{1}{7}$- $\dfrac{1}{9}$ + … + (-$\dfrac{1}{2019}$) + $\dfrac{1}{2019}$ – $\dfrac{1}{2021}$ ]
= $\dfrac{1}{2}$ . [ 1 – $\dfrac{1}{2021}$]
= $\dfrac{1}{2}$ . [ $\dfrac{2021}{2021}$ -$\dfrac{1}{2021}$ ]
= $\dfrac{1}{2}$ . $\dfrac{2020}{2021}$
=$\dfrac{1010}{2021}$
Đề thiếu nha bạn.
$\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + …. + \dfrac{1}{2019.2021}$
$= \dfrac{1}{2} . (\dfrac{2}{1.3} + \dfrac{2}{3.5} + \dfrac{2}{5.7} + …. + \dfrac{2}{2019.2021})$
$= \dfrac{1}{2} . (1 – \dfrac{1}{2021})$
$= \dfrac{1}{2} . \dfrac{2020}{2021}$
$= \dfrac{1010}{2021}$