Question

\(B=\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+....+\frac{9}{77.80}\)

Answer 1

Ta có B= 3(\(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\) )

         B= 3\(\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)

        B= 3.(\(\frac{1}{20}-\frac{1}{80}\))

        B=3.\(\frac{3}{80}\)=\(\frac{9}{80}\)

Answer 2

\(\frac{3}{9}B=\frac{3}{9}.\left(\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+...+\frac{9}{77.80}\right)\)

=> \(\frac{3}{9}B=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+....+\frac{3}{77.80}\)

=\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+....+\frac{1}{77}-\frac{1}{80}\)

=\(\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)

=> \(B=\frac{3}{80}:\frac{3}{9}=\frac{3}{80}.\frac{9}{3}=\frac{9}{80}\)