\(A=\frac{1}{1.300}+\frac{1}{2.301}+..........+\frac{1}{101.400}\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+........+\frac{299}{101.400}\)

\(\Rightarrow299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...........+\frac{1}{101}-\frac{1}{400}\Rightarrow299A=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+.......+\frac{1}{400}\right)\)\(\Rightarrow\)\(A=\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)

là sao

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

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

cái lịt

có học thì mới có ăn không làm mà đòi có điểm chỉ có ăn ...

cghsbbvb hs bsc x bvbddddddd c n  snsnfERGQHZ NAC nnnnNNNNNNNNNNNNNNNNNNNNNNNN nsn  v 

tgrtyfdytiloniyu7d tadftr DxZhfhygd ỳdstAACA 

ngu vậy chết mm đi