\(C=\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\)

\(\Rightarrow\frac{2}{3}C=\frac{2}{3}\cdot\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\right)\)

\(\Rightarrow\frac{2}{3}C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{47\cdot49}\)

\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)

\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{49}\)

\(\Rightarrow\frac{2}{3}C=\frac{46}{147}\)

\(\Rightarrow C=\frac{46}{147}:\frac{2}{3}\)

\(\Rightarrow C=\frac{23}{49}\)

3/3.5+3/5.7+3/7.9+.....+3/47.49

=1-1/5+1/5-1/7+...+1/47-1/49

=1-1/49

=48/49

C = 3/2 . (2/3.5 + 2/5.7 + 2/7.9 + ... + 2/47.49)

C = 3/2 . (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/47 - 1/49)

C = 3/2 . (1/3 - 1/49) = 3/2 . 46/147 = 23/49

Chúc bn HỌC TỐT nhá!!!!

=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

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

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

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

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

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

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

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

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

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

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

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

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

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

NHẦM GIẢI LẠI :

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

a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)

\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

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

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)

b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(=\frac{15}{16}\)

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!1111

\(a)\) \(A=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{49.51}\)

\(A=3\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.50}\right)\)

\(2A=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.50}\right)\)

\(2A-A=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=3\left(\frac{1}{3}-\frac{1}{50}\right)\)

\(A=1-\frac{3}{50}\)

\(A=\frac{47}{50}\)

Vậy \(A=\frac{47}{50}\)

\(b)\) \(B=\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(B=1-\frac{1}{16}\)

\(B=\frac{15}{16}\)

Vậy \(B=\frac{15}{16}\)

Chúc bạn học tốt ~

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a