A = \(\dfrac{2}{35}\) + \(\dfrac{4}{77}\) + \(\dfrac{2}{143}\) + \(\dfrac{4}{221}\) + \(\dfrac{2}{323}\) + \(\dfrac{4}{437}\) + \(\dfrac{2}{575}\)

A =  \(\dfrac{2}{5\times7}\)+\(\dfrac{4}{7\times11}\)+\(\dfrac{2}{11\times13}\)+\(\dfrac{4}{13\times17}\)+\(\dfrac{2}{17\times19}\)+\(\dfrac{4}{19\times23}\)+\(\dfrac{2}{23\times25}\)

A = \(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)-\(\dfrac{1}{19}\)+\(\dfrac{1}{19}\)-\(\dfrac{1}{23}\)+\(\dfrac{1}{23}\)-\(\dfrac{1}{25}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{25}\)

A = \(\dfrac{4}{25}\)

\(Q=\frac{5}{30}+\frac{5}{30}+\frac{5}{30}+\frac{5}{30}+\frac{5}{30}\)

\(=\frac{5}{30}.5\)

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

5/30+15/90+25/150+35/210+45/270

=1/6+1/6+1/6+1/6+1/6

=1/30

11,6

:v 

\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+\dfrac{61}{63}+\dfrac{97}{99}\)

\(=\left(1-\dfrac{2}{3}\right)+\left(1-\dfrac{2}{15}\right)+\left(1-\dfrac{2}{35}\right)+\left(1-\dfrac{2}{63}\right)+\left(1-\dfrac{2}{99}\right)\)

\(=\left(1+1+1+1+\right)-\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\)

\(=5-\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)\)

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

\(=5-\left(1-\dfrac{1}{11}\right)\)

\(=5-\dfrac{10}{11}\)

\(=\dfrac{45}{11}\)

A = \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)

A = \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{99\times100}\)

A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A = \(\dfrac{1}{3}\) - \(\dfrac{1}{100}\)

A = \(\dfrac{97}{300}\) 

Lời giải:

Gọi tổng trên là $A$

$A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{99.100}$

$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}$

$=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}$

\(B=\frac{5}{30}+\frac{15}{90}+\frac{25}{150}+\frac{35}{210}+\frac{45}{270}\)

\(B=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{5}{6}\)

\(A=29\dfrac{1}{2}\cdot\dfrac{2}{3}+39\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{2}\cdot\dfrac{2}{3}+\dfrac{118}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{118}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{59}{2}+\dfrac{5}{6}\)

\(=59\cdot\left(\dfrac{1}{3}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=\dfrac{5}{6}\cdot\left(59+1\right)=\dfrac{5}{6}\cdot60=50\)