^{\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{57.59}\)}

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{59}\)

\(B=1-\frac{1}{59}\)

\(B=\frac{59}{59}-\frac{1}{59}=\frac{58}{59}\)

Vậy B = \(\frac{58}{59}\)

Lưu ý: Dấu "." là dấu nhân

\(B=\frac{2}{1.3}+\frac{1}{3.5}+\frac{2}{5.7}+...+\frac{1}{57.59}\)

\(B=1.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{57}-\frac{5}{59}\right)\)

\(B=1.\left(1-\frac{1}{59}\right)\)

\(B=1.\frac{58}{59}\)

\(B=\frac{58}{59}\)

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\frac{10}{11}\)

\(=\frac{5}{11}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\frac{10}{11}\)

\(=\frac{5}{11}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)

\(B=5\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{100}-\frac{1}{103}\right)\)

\(\Rightarrow B=5\cdot\left(1-\frac{1}{103}\right)=5\cdot\frac{102}{103}=\frac{510}{103}\)

\(C=5\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}\right)\)

\(\Rightarrow C=5\cdot\left(1-\frac{1}{101}\right)=5\cdot\frac{100}{101}=\frac{500}{101}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(B=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(B=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(C=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(C=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(C=\frac{5}{2}\left(1-\frac{1}{101}\right)\)

\(C=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)

\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(A=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

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

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

# Học tốt #

ngày mai mik làm đc ko

ok ai giải được giúp mik nha chiều mai mik phải nộp rồi

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

\(S.2=\frac{1}{1}-\frac{1}{11}\)

\(S.2=\frac{10}{11}\)

\(S=\frac{10}{11}:2\)

\(S=\frac{5}{11}\)

S = 5/11

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(=\frac{1}{1}-\frac{1}{11}\)

\(=\frac{10}{11}\)

= 1-1/3 + 1/3-1/5+.......+1/97-1/99

=  1 - 1/99

= 98/99

sao lại là 1- 1/3 + 1/3 -1/5 + ...... 1/97 - 1/99 hả bạn :|

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{95.97}\)\(+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{13.15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

mik đã trả lời rồi mà , sao chưa hiện ra ????

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}+\frac{2}{13\times15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}=\frac{14}{15}\)

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

\(y=\frac{2}{3}.\frac{11}{10}\)

\(y=\frac{22}{30}\)

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

         \(\frac{10}{11}.y=\frac{2}{3}\)

                    \(y=\frac{11}{15}\)

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{11-9}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

         \(y=\frac{2}{3}:\frac{10}{11}\)

         \(y=\frac{11}{15}\)