Ta có :
\(\frac{1}{51}\)> \(\frac{1}{100}\)
\(\frac{1}{52}\)> \(\frac{1}{100}\)
...
\(\frac{1}{99}\)> \(\frac{1}{100}\)
\(\frac{1}{100}\)= \(\frac{1}{100}\)
=> S > 50 x \(\frac{1}{100}\)
=> S > \(\frac{50}{100}\)= \(\frac{1}{2}\)
Vậy S > \(\frac{1}{2}\)
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Ta có \(\frac{1}{51}>\frac{1}{100}\)
\(\frac{1}{52}>\frac{1}{100}\)
...
\(\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)
( có 50 phân số)
\(\Rightarrow S>50.\frac{1}{100}\)
\(\Rightarrow S>\frac{1}{2}\)
Vậy...
Bài làm
Ta thấy: \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}+\frac{1}{100}\)có 50 số hạng
=> \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}\)có 49 số hạng
Và \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}\)luôn lớn hơn \(\frac{1}{100}\)
Ta có: \(\frac{1}{100}\Rightarrow\frac{1}{100}.50=\frac{50}{100}=\frac{1}{2}\)
=> \(\frac{1}{100}=\frac{1}{2}\)
=> \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}>\frac{1}{2}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{2}\)
Vậy S > 1/2
# Học tốt #
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