S= 1/199 + 2/198 + ... + 198/2 + 199/1

S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2  + 1) +1

S= 200/200 + 200/199 + 200/198 + ... + 200/2 

S= 200.(1/200 + 1/199 + ... + 1/2)

Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)

B=1 : 200 = 1/200

2/

a, Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2005^2}< \frac{1}{2004.2005}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}=B\)

b, \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}=1-\frac{1}{2005}< 1\)

3/ 

Ta có: \(\frac{1}{101}< \frac{1}{100};\frac{1}{102}< \frac{1}{100};...;\frac{1}{200}< \frac{1}{100}\)

\(\Rightarrow A< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\left(100ps\right)=\frac{1}{100}\cdot100=1\left(1\right)\)

Lại có: \(\frac{1}{101}>\frac{1}{200};\frac{1}{102}>\frac{1}{200};...;\frac{1}{200}=\frac{1}{200}\)

\(\Rightarrow A>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\left(100ps\right)=\frac{1}{200}\cdot100=\frac{1}{2}\left(2\right)\)

Từ (1) và (2) => \(\frac{1}{2}< A< 1\)

vận dụng 3 A nha bn

dễ mà

xong tìm A ok nha bn

ok

1/

\(A=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)

=\(\frac{1}{101}\left(\frac{101}{1.102}+\frac{101}{2.103}+...+\frac{101}{299.400}\right)\)

=\(\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)

=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+...+\frac{1}{400}\right)\right]\)

=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)+\left(\frac{1}{102}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+...+\frac{1}{299}\right)-\left(\frac{1}{300}+...+\frac{1}{401}\right)\right]\)

=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{401}\right)\right]\)

Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)

\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)

\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)

\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)

Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)

\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)

\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)

\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)

1/

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)

\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)

\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

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

\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)

2/ 

\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)

\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)

\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

câu 1 bảo binfhd dúng nhé

\(A=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.......+\frac{1}{101}-\frac{1}{400}\right)\)

\(A=\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+....+\frac{1}{400}\right)\right]\)

=>đpcm

\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{.302}+....+\frac{1}{101.400}\)

=> \(299.A=\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}\)

=> \(299.A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)

=> \(A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

Có j không hiểu có thể hỏi lại mk

Chúc bạn làm bài tốt

\(A=\frac{1}{1\times300}+\frac{1}{2\times301}+\frac{1}{3\times302}+...+\frac{1}{100\times399}+\frac{1}{101\times400}\)

\(A=\frac{1}{299}\times\left(\frac{299}{1\times300}+\frac{299}{2\times301}+\frac{299}{3\times302}+...+\frac{299}{100\times399}+\frac{299}{101\times400}\right)\)

\(A=\frac{1}{299}\times\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{100}-\frac{1}{399}+\frac{1}{101}-\frac{1}{400}\right)\)

\(A=\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{399}+\frac{1}{400}\right)\right]\)

=> đpcm