Question

A = \(\frac{3}{5\cdot11}+\frac{5}{11\cdot21}+\frac{7}{21\cdot35}+\frac{9}{35\cdot53}\) =?

 

Answer 1

\(\frac{3}{5.11}+\frac{5}{11.21}+\frac{7}{21.35}+\frac{9}{35.53}=\frac{1}{2}\left(\frac{6}{5.11}+\frac{10}{11.21}+\frac{14}{21.35}+\frac{18}{35.53}\right)\)

\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{2}.\frac{48}{265}=\frac{24}{265}\)

Answer 2

Sửa lại :

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{2}.\frac{48}{265}=\frac{24}{265}\)

 

Answer 3

\(2.A=\frac{6}{5.11}+\frac{10}{11.21}+\frac{14}{21.35}+\frac{18}{35.53}\)

\(2.A=\frac{11-5}{5.11}+\frac{21-10}{11.21}+\frac{35-21}{21.35}+\frac{53-35}{35.53}=\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\)

\(2A=\frac{1}{5}-\frac{1}{53}=\frac{48}{265}\)

A = \(\frac{48}{265}:2=\frac{24}{265}\)