\(S=7(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{63}) \)

\(S=7(\frac{1}{3}-\frac{1}{63})\)

\(S=7(\frac{21}{63}-\frac{1}{63}) \)

\(S=7.\frac{20}{63}\)

\(S=\frac{20}{9}\)

Do đó:\(S<\frac{5}{2}\)

S=\(\frac{2.7}{3.5}+\frac{2.7}{5.7}+\frac{2.7}{7.9}+....+\frac{2.7}{61.63}\)và\(\frac{5}{2}\)

S=7.(\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....-\frac{1}{63}\)) và\(\frac{5}{2}\)

S=7.(\(\frac{1}{3}-\frac{1}{63}\)) và\(\frac{5}{2}\)

S=7.\(\frac{20}{63}\)và\(\frac{5}{2}\)

=>S=\(\frac{20}{9}\)so với \(\frac{5}{2}\)

=>S=\(\frac{40}{18}\)và\(\frac{45}{18}\)

=>S<\(\frac{5}{2}\)

M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))

M= 3.(\(\frac{1}{5}-\frac{1}{61}\))

M=\(\frac{168}{305}\)

\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(M=\frac{84}{305}\)

\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Chúc bạn học tốt ~ 

làm hộ mk lun câu b ik

a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-(1-\frac{1}{199})\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Vậy \(A=\frac{-197}{199}\)

b) cũng thế 

BÀi này dễ thôi bạn ạ 

N=3(1/5.7+1/7.9+.........+1/197.199)

N=3/2( 1/5-1/7+1/7-1/9+1/9-..........+1/197-1/199)

N=3/2(1/5-1/199)

N=3/2.194/995

N=291/995

k đúng cho mình nhé

N=3.1/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/197-1/199)

N=3.1/2.(1/5-1/99)

N=3.1/2.94/495

N=3.47/495

N=47/165

N=3/5.7+3/7.9+........+3/197.199

=>N=3/2(2/5.7+2/7.9+.....+2/197.199)

=>N=3/2(1/5-1/7+1/7-1/9+........+1/197-1/199)

=>N=3/2(1/5-1/199)

=>N=3/2x194/995=291/995

A. Đặt A= biểu thức đã cho

=>\(\frac{A}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>\(\frac{A}{3}.2=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

=>\(\frac{2A}{3}-\frac{A}{3}=2-\frac{1}{2^9}\)

=>\(A=\frac{3\left(2^{10}-1\right)}{2^9}\)

B. Đặt B=biểu thức đã cho

\(\Rightarrow\frac{B}{2}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(=\frac{1}{3}-\frac{1}{2017}=\frac{2014}{6051}\)

\(\Rightarrow B=\frac{4028}{6051}\)

=3(1/1.3+1/3.5+1/5.7+1/7.9)

=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2

=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0

=3/2.8/9=4/3

ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)

ta đặt 3 làm chung rồi tự làm đc

dat 3 lam chung

\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)

\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)

\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)

\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(=4\cdot\frac{100}{309}=\frac{400}{309}\)

NHẦM GIẢI LẠI :

\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)