Biến đổi vp của đẳng thức :

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right]\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)

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ngu vảy 

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{100}\right)\)

\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(\Leftrightarrow\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Ta có đpcm

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{100}-1-\frac{1}{2}-...-\frac{1}{50}\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

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~

Ta có:\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+............+\frac{1}{100}\)

\(=\left(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+........+\frac{1}{100}\right)\)

\(>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{1}{2}\)

\(\left(\frac{1}{51}+\frac{1}{52}+..........+\frac{1}{75}\right)+\left(\frac{1}{76}+........+\frac{1}{100}\right)\)

\(< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}< 1\)

\(\Rightarrowđpcm\)

Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)

Vì mọi phân số của tổng đều nhỏ hơn 1 nên tổng đó nhỏ hơn 1.

k nha

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrowđpcm\)

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