Sửa đề: \(A=\frac{4}{3\cdot5}-\frac{6}{5\cdot7}+\frac{8}{7\cdot9}-\frac{10}{9\cdot11}+\cdots+\frac{100}{99\cdot101}\)

\(=\frac12\left(\frac{8}{3\cdot5}-\frac{12}{5\cdot7}+\frac{16}{7\cdot9}-\frac{20}{9\cdot11}+\cdots+\frac{200}{99\cdot101}\right)\)

\(=\frac12\left(\frac13+\frac15-\frac15-\frac17+\cdots+\frac{1}{99}+\frac{1}{101}\right)\)

\(=\frac12\left(\frac13+\frac{1}{101}\right)=\frac12\cdot\frac{104}{303}=\frac{52}{303}\)

dễ thế mà mày ko bít làm à

B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101) 

B=2(1/3.5 -2/5.7+..............+1/99.100)

B=2(1/3-1/5+1/5-.............+1/99-1/100)

B=2(1/3-1/100)

B=2.97/100

B=97/50

\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)

\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)

\(2x+\frac{8}{33}=-\frac{2016}{2017}\)

\(2x=\frac{-2016}{2017}-\frac{8}{33}\)

\(2x=\frac{-2024}{2017}\)

\(x=-\frac{1012}{2017}\)

\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)

\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)

\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)

\(2x+\frac{8}{33}=\frac{-2016}{2017}\)

\(2x=\frac{-2016}{2017}-\frac{8}{33}\)

Số dư dài quá. Đến đây bạn tự làm tiếp nhé

a.  

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

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn