\(\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2021}{2020.2021}-\dfrac{2020}{2020.2021}=\dfrac{2021-2020}{2020.2021}=\dfrac{1}{2020.2021}\)
\(\dfrac{1}{2020\cdot2021}=\dfrac{2021-2020}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)(đpcm)
\(\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2021}{2020.2021}-\dfrac{2020}{2020.2021}=\dfrac{2021-2020}{2020.2021}=\dfrac{1}{2020.2021}\)
\(\dfrac{1}{2020\cdot2021}=\dfrac{2021-2020}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)(đpcm)
Chứng minh rằng
\(\dfrac{5}{13.18}=\dfrac{1}{13}-\dfrac{1}{8}\)
chứng minh rằng
\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\) và B= 2
Chứng minh rằng
\(\dfrac{1}{n.\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\left(nEN^{\cdot}\right)\)
Chứng minh rằng :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}>\dfrac{25}{12}\)
Chứng minh rằng
\(\dfrac{k}{n.\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\left(n;kEN^{\cdot}\right)\)
mấy bạn ơi giúp mình câu này với
chứng minh rằng: \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{4010^2}< \dfrac{1}{2}\)
chứng tỏ rằng:
E=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)<\(\dfrac{3}{4}\)
so sanh
a)-2021/2020 va 1/2
b)2022/2021 va 2021/2020
Cho B=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{2014}{5^{2015}}\). Chứng tỏ rằng B<\(\dfrac{1}{16}\)