\(\dfrac{1}{13}-\dfrac{1}{18}=\dfrac{18}{13.18}-\dfrac{13}{13.18}=\dfrac{18-13}{13.18}=\dfrac{5}{13.18}\)
\(\dfrac{5}{13\cdot18}=\dfrac{18}{13\cdot18}-\dfrac{13}{13\cdot18}=\dfrac{1}{13}-\dfrac{1}{18}\)(đpcm)
\(\dfrac{1}{13}-\dfrac{1}{18}=\dfrac{18}{13.18}-\dfrac{13}{13.18}=\dfrac{18-13}{13.18}=\dfrac{5}{13.18}\)
\(\dfrac{5}{13\cdot18}=\dfrac{18}{13\cdot18}-\dfrac{13}{13\cdot18}=\dfrac{1}{13}-\dfrac{1}{18}\)(đpcm)
chứng minh rằng
\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\) và B= 2
Chứng minh rằng :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}>\dfrac{25}{12}\)
Chứng minh rằng
\(\dfrac{1}{2020.2021}=\dfrac{1}{2020-2021}\)
Chứng minh rằng
\(\dfrac{1}{n.\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\left(nEN^{\cdot}\right)\)
Chứng minh rằng
\(\dfrac{k}{n.\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\left(n;kEN^{\cdot}\right)\)
Cho B=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{2014}{5^{2015}}\). Chứng tỏ rằng B<\(\dfrac{1}{16}\)
ko quy đồng hãy so sánh
a) A=\(\dfrac{3}{8^3}\)+\(\dfrac{7}{8^4}\)
B=\(\dfrac{7}{8^3}\)+\(\dfrac{4}{8^4}\)
b)A= \(\dfrac{5^{12}+1}{5^{13}+1}\)
B=\(\dfrac{5^{11}+1}{5^{12}+1}\)
1. Chứng tỏ rằng:
a) \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)
b) \(\dfrac{m}{a.\left(a+m\right)}=\dfrac{1}{a}-\dfrac{1}{a+m}\)
2. Tính
a) \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
b) \(\dfrac{5}{10.15}+\dfrac{5}{15.20}+...+\dfrac{5}{195.200}\)
c) \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{96.98}\)
mấy bạn ơi giúp mình câu này với
chứng minh rằng: \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{4010^2}< \dfrac{1}{2}\)