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

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

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

a=4/3.7 +4/7.11+4/11.15 +.....+4/107/111

=1/3-1/7+1/7-1/11+1/11-1/15+......+1/107-1/111

=1/3-1/111

=12/37

12/37nhe

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

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

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

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

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

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

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

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

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

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

lớp 6

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

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 -_-

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

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

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

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

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

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

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

Ta có:

\(C=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{40.41}+\frac{2}{41.42}\)

\(\Rightarrow C=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}+\frac{1}{41.42}\right)\)

\(\Rightarrow C=2\left(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{41-40}{40.41}+\frac{42-41}{41.42}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}+\frac{1}{41}-\frac{1}{42}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{42}\right)=\frac{13}{21}\)

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

\(\Rightarrow D=\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{111-107}{107.111}\)

\(\Rightarrow D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)\(E=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

\(\Rightarrow E=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}+\frac{11-10}{10.11}\)

\(\Rightarrow E=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)