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

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\) \(=\dfrac{5}{39}\)

\(\dfrac{5}{39}\)

=\(\dfrac{5}{39}\)

\(2A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}=\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=\dfrac{1}{3}-\dfrac{1}{13}=\dfrac{13-3}{39}=\dfrac{10}{39}\)

\(\Rightarrow A=\dfrac{10}{39}.\dfrac{1}{2}=\dfrac{10}{78}=\dfrac{5}{39}\)

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

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

=1/2*10/39

=5/39

2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)

\(B=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

=> \(2B=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\) => \(2B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\) => \(2B=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+\dfrac{11-9}{9.11}+\dfrac{13-11}{11.13}\) => \(2B=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+\dfrac{11}{9.11}-\dfrac{9}{9.11}+\dfrac{13}{11.13}-\dfrac{11}{11.13}\) => \(2B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\) => \(2B=\dfrac{1}{3}-\dfrac{1}{13}\)

=> \(B=\left(\dfrac{13}{39}-\dfrac{3}{39}\right):2\)

=> \(B=\dfrac{10}{39}.\dfrac{1}{2}\)

=> \(B=\dfrac{10}{39.2}\)

=> \(B=\dfrac{5}{39}\)

Vậy \(B=\dfrac{5}{39}\)

\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+\dfrac{11-9}{9.11}+\dfrac{13-11}{11.13}\right)\)

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

\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

\(B=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)

\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{11.13}\)

Ta có công thức:

\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

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

\(B=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

\(B=\dfrac{1}{2}.\left(\dfrac{10}{39}\right)=\dfrac{5}{39}\)

MINK NGHĨ VẬY NHA !!

CHÚC BẠN HỌC TỐT !!!!!

I. Tính:

1) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(=1-\dfrac{1}{6}\)

\(=\dfrac{5}{6}\)

2) \(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)

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

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

\(=\dfrac{1}{3}-\dfrac{1}{13}\)

\(=\dfrac{13}{39}-\dfrac{3}{39}=\dfrac{10}{39}\)

II. Tìm x:

\(1\dfrac{3}{5}+\left(\dfrac{\dfrac{2}{171}}{\dfrac{5}{171}}+\dfrac{\dfrac{2}{373}}{\dfrac{5}{373}}\right)x=\dfrac{16}{5}\)

\(\dfrac{8}{5}+\left[\dfrac{2\left(\dfrac{1}{171}+\dfrac{1}{373}\right)}{5\left(\dfrac{1}{171}+\dfrac{1}{373}\right)}\right]x=\dfrac{16}{5}\)

\(\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\)

\(\dfrac{2}{5}x=\dfrac{16}{5}-\dfrac{8}{5}\)

\(\dfrac{2}{5}x=\dfrac{8}{5}\)

\(x=\dfrac{8}{5}:\dfrac{2}{5}\)

\(x=4\)

`1/15+1/35+1/63+1/99+1/143`

`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`

`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`

`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`

`=1/2.(1/3-1/13)`

`=1/2 . 10/39`

`=5/39`