Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

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

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Câu 1:

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(\dfrac{1}{3}-\dfrac{1}{101}\)

= \(\dfrac{98}{303}\)

Câu 2 làm tương tự ở câu 1 nhé

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2020.2022}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}\)

\(=\dfrac{2021}{2022}\)

2/2*[2/1-2/2022]=2021/1011

\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}\)

\(=\dfrac{133}{195}\)

=1-(1/3-1/5+1/5-1/7+...+1/61-1/63)

=1-20/63=43/63

\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.100}\)

\(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99.100}+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{3}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(B=\dfrac{1}{3}-\dfrac{2}{100}+\dfrac{1}{99}\)

\(B=\dfrac{1}{3}-\dfrac{1}{50}+\dfrac{1}{99}\)

Đến đây thì hết tính hợp lý được rồi:v

\(B=\dfrac{34}{99}-\dfrac{1}{50}\)

\(B=\dfrac{1601}{4950}\)

101 chứ bạn

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)

= 2/3 . 2/5 + 2/5 . 2/7 + ... + 2/99 . 2/101

= 2/3 - 2/5 + 2/5 - 2/7 + ... + 2/99 - 2/101

= 2/3 - 2/101

= 196/303

2/3 - 2/5 + 2/5 - 2/7 + 2/7 - 2/9 + .... + 2/97 - 2/99 + 2/99 - 2/101

 = 2/3 - 2/101

= 196/303

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{39}\)

\(=(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})+\dfrac{2}{39}\)

\(=(\dfrac{1}{3}-\dfrac{1}{13})+\dfrac{2}{39}\)

\(=\dfrac{10}{39}+\dfrac{2}{39}\)

\(=\dfrac{4}{13}\)

gọi biểu thức đó là A

A=\(1.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)+\dfrac{2}{39}\)

A= \(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)+\dfrac{2}{39}=\dfrac{4}{13}\)

mk nhanh nhất nha bạn

\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)

\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(M=2.\dfrac{32}{99}\)

\(M=\dfrac{64}{99}\)

http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp

\(m=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

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

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)

Chúc bạn học tốt!!!

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

= \(\dfrac{1}{3}-\dfrac{1}{15}\)

= \(\dfrac{4}{15}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

\(S=\dfrac{5-3}{5.3}+\dfrac{7-5}{7.5}....+\dfrac{25-23}{23.25}\)

\(S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{23}-\dfrac{1}{25}\)

\(S=\dfrac{1}{3}-\dfrac{1}{25}=\dfrac{25-3}{3.25}=\dfrac{7}{25}\)

sửa lại nha bạn

\(\dfrac{25-3}{25.3}=\dfrac{22}{75}\)

\(S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{23.25}\)

\(S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{23}-\dfrac{2}{25}\)

\(S=\dfrac{1}{3}-\dfrac{1}{25}\)

\(S=\dfrac{25}{75}-\dfrac{3}{75}\)

\(S=\dfrac{22}{75}\)