A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)

=3/2x4/3x...............x100/99

=2-1/99

=197/99

A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)

A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)

A=\(\frac{100}{2}=50\)

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%

Câu đầu tiên:

\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)

\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}=\frac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{3\cdot4\cdot5\cdot...\cdot99\cdot2}=\frac{100}{2}=50\)

Câu thứ 2:

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}>\frac{32}{100}\)

N ở đâu hả bạn

N là số tự nhiên đó

a. \(A=\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(\Rightarrow A=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow A=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow A=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(\Rightarrow A=2.\frac{32}{99}\)

\(\Rightarrow A=\frac{64}{99}\)

b. \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\notin N\)

Ta cần chứng minh M < 1 và M khác 0

Dễ thấy M khác 0 ( 1 )

Ta có : \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(M< 1-\frac{1}{2020}=\frac{2019}{2020}\)

=> M < 1 ( 2 )

Từ ( 1 ) và ( 2 ) => Đpcm

A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được: 

\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)

\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)

=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)

\(4A>1=>A>\frac{1}{4}\)

Bn trừ 2 PS kiểu gì hay zậy? 

Giúp mình nhá

Ta có : 

\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)

\(=\)\(2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

\(=\)\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\)\(2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\)\(2.\frac{32}{99}\)

\(=\)\(\frac{64}{99}\)

Chúc bạn học tốt ~ 

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

Tự tính

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

32/99

k với nghe bạn

và chúc chueeuf nay thi tốt

\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right).\)

 \(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}x\frac{32}{99}=\frac{32}{198}\)

bn tự rút gọn nha mk mới làm tắt đó

Ta có : \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{33}{99}-\frac{1}{99}\)

\(=\frac{32}{99}\)

32/99

a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

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

\(=4.\frac{32}{99}\)

\(=\frac{128}{99}\)

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}\)

\(=\frac{64}{99}\)

\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

\(=6\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{206}\right)\)

\(=6.\frac{102}{206}\)

\(=\frac{612}{206}\) ( tự rút gọn