ta có 1/51>1/100
1/52>1/100
..................
1/100=1/100
\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2
\(\Rightarrow\)S>\(\frac{1}{2}\)
cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá
chúc bạn học tốt~
\(ChoS=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.....+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\)
HÃY SO SÁNH \(S\)VỚI \(\frac{1}{2}\)
Cho S =\(\frac{1}{50}\)+\(\frac{1}{51 }\)+\(\frac{1}{52}\)+...+\(\frac{1}{98}\)+\(\frac{1}{99}\)
Chứng tỏ rằng S >\(\frac{1}{2}\)
DDODOGDOGE
Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
Chứng minh rằng :\(\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Chứng minh rằng
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Tính: \(B=\frac{100^2+1^2}{100\cdot1}+\frac{99^2+2^2}{99\cdot2}+\frac{98^2+3^2}{98\cdot3}+...+\frac{52^2+49^2}{52\cdot49}+\frac{51^2+50^2}{51\cdot50}\)
Chứng tỏ rằng: \(\frac{1}{1}\times\frac{1}{3}\times\frac{1}{5}\times.....\times\frac{1}{99}=\frac{2}{51}\times\frac{2}{52}\times\frac{2}{53}\times.....\times\frac{2}{100}\)
Tính:\(\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\right):\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\right)\)
Chứng minh rằng:
(1+\(\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\))-(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\))=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
