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)

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 :

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

\(\Rightarrow H>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\)

\(\Rightarrow H>\frac{1}{100}.50\)

\(\Rightarrow H>\frac{1}{2}\)

Lại có :

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

\(\Rightarrow H< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+........+\frac{1}{51}\)

\(\Rightarrow H< \frac{1}{51}.50\)

\(\Rightarrow H< \frac{50}{51}\)

\(\Rightarrow H< 1\)

Vậy \(\frac{1}{2}< H< 1\left(ĐPCM\right)\)

Ta có :

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

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

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

cảm ơn bạn nha

ai giúp mk ik

mk đg cần gấp,còn nhìu đề chx lm

đánh phần ở đâu thế?

Số chia rút gọn thành 1/51+1/52+...+1/99+1/100

=> biểu thức bằng 1

\(A=\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Chia A làm 2 phần,mỗi phân 25 số hạng.

\(A>\frac{25.1}{75}+\frac{25.1}{100}\)

\(A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

Bé hơn em làm tương tự có điều để nguyên cả 50 p/số.

Chúc em học tốt^^

bạn có thể giải cụ thể hơn cho mình được ko ?

mình chả hiểu gì cả

ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

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

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

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