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Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
S = 1 / 50 + 1 / 51 +...+ 1 / 99 > 1 / 99 + 1 / 99 +...+ 1 / 99 = 50 / 99 > 50 / 100 = 1/2
\(S=\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(có 50 số hạng)\(=\frac{50}{100}=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\) .
Có: \(\frac{1}{50}>\frac{1}{100}\\ \frac{1}{51}>\frac{1}{100}\\ \frac{1}{52}>\frac{1}{100}\\ .\\ .\\ .\\ \frac{1}{98}>\frac{1}{100}\\ \frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)(có 50 số hạng \(\frac{1}{100}\))
\(\Rightarrow S>\frac{1}{100}\cdot50\)
\(\Rightarrow S>\frac{50}{100}\)
\(\Rightarrow S>\frac{1}{2}\left(đpcm\right)\)
\(S=\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{99}>\frac{1}{99}+\frac{1}{99}+...+\frac{1}{99}=\frac{50}{99}>\frac{50}{100}=\frac{1}{2}\)
Ta có:
\(\frac{1}{2}=\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(50 PS)
Vì \(\frac{1}{50}>\frac{1}{100}\)
\(\frac{1}{51}>\frac{1}{100}\)
\(\frac{1}{52}>\frac{1}{100}\)
.......................
\(\frac{1}{99}>\frac{1}{100}\)
\(=>\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\right)>\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\) ( có 50 PS)
\(=>\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}>\frac{1}{2}\)
\(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\Rightarrow M< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow M< 1-\frac{1}{99}< 1\)
Dễ thấy M > 0 nên 0 < M < 1
Vậy M không là số tự nhiên.
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
\(\Rightarrow S>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\) (50 số hạng \(\frac{1}{100}\))
\(\Rightarrow S>\frac{1}{100}.50=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\left(đpcm\right)\)
Ta có: \(\frac{1}{50}\)>\(\frac{1}{100}\)
\(\frac{1}{51}\)>\(\frac{1}{100}\)
\(\frac{1}{52}\)>\(\frac{1}{100}\)
..................
\(\frac{1}{99}\)>\(\frac{1}{100}\)
=>\(\frac{1}{50}\)+\(\frac{1}{51}\)+.............+\(\frac{1}{99}\)>\(\frac{1}{100}\).50=\(\frac{1}{2}\)(50 là số số hạng của S nha)
=>S>\(\frac{1}{2}\)