\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(\Rightarrow=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{12}{37}\)

k nha

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{108}{333}=\frac{12}{37}\)

ten gia mao nobita 

nobi k the gioi nhu the 

do gia :v

4x(\(\frac{1}{3.7}+...+\frac{1}{107.111}\) )

4(\(\frac{1}{3}-\frac{1}{7}+...+\frac{1}{107}-\frac{1}{111}\))

4(\(\frac{1}{3}-\frac{1}{111}\))

4.\(\frac{12}{37}\)

48/37

Ta có A = \(\frac{4}{3.7}+\frac{4}{7.11}+..............+\frac{4}{107.111}\)

=> A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.............+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

k nha bạn

\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)

\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)

\(A=4.\frac{12}{37}\)

\(A=\frac{48}{37}\)

lớp 6

a.\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}=\frac{37}{111}-\frac{1}{111}=\frac{36}{111}=\frac{12}{37}\)

Vậy A=\(\frac{12}{37}\)

b.\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)

Vậy \(B=\frac{2}{7}\)

c.\(C=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

\(\Rightarrow C.\frac{1}{2}=\left(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\right).\frac{1}{2}\)

\(=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)

\(=\frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}\)

Vậy \(C=\frac{3}{16}\)

A = \(\frac{4}{3.7}+\frac{4}{7.9}+...+\frac{4}{107.111}\)

A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}\)=\(\frac{12}{37}\)

2 câu sau tương tự. Mik ngại làm lắm -_-

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)

Đặt A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{103.107}+\frac{4}{107.111}\)

\(=\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right)+...+\left(\frac{1}{103}-\frac{1}{107}\right)+\left(\frac{1}{107}-\frac{1}{111}\right)\)

\(=\frac{1}{3}-\frac{1}{111}=\frac{37-1}{111}=\frac{36}{111}\)

Viết vậy đúng đó em

A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069

\(B=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}\left(\frac{1}{7}-\frac{1}{11}\right)+....+\frac{1}{4}\left(\frac{1}{107}-\frac{1}{111}\right)\)

  \(=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{107}-\frac{1}{111}\right)\)

  \(=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{111}\right)\)

  \(=\frac{107}{444}\)

vậy biểu thức trên =\(\frac{107}{444}\)

xong zùi đó bn ak

B=\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(B=\frac{1}{3}-\frac{1}{111}\)

\(B=\frac{111}{333}-\frac{3}{333}\)

\(B=\frac{108}{333}\)

\(B=\frac{12}{37}\)

Vậy B =\(\frac{12}{37}\)