\(\frac{1}{5}+\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}+\frac{1}{230}+\frac{1}{299}\)

\(=\frac{1}{1.5}+\frac{1}{5.4}+\frac{1}{4.11}+\frac{1}{11.7}+...+\frac{1}{23.13}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{4}+...+\frac{1}{23}-\frac{1}{13}\)

\(=1-\frac{1}{13}\)

\(=\frac{12}{13}\)

Đặt: \(\frac{1}{117}=a,\frac{1}{119}=b\)

Khi đó: \(A=3ab-4a.5.118b-5ab+\frac{8}{39}\)

\(=-2362ab+\frac{8}{39}\)

\(=-2362.\frac{1}{117}.\frac{1}{119}=\frac{38}{1071}\)

mik ko hieu lam

=10/117

bạn giải ra đi

Câu C giải rồi

\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)

\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)

\(\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)

=\(\dfrac{2}{10}+\dfrac{2}{40}+\dfrac{2}{88}+\dfrac{2}{154}+\dfrac{2}{238}+\dfrac{2}{340}+\dfrac{2}{460}+\dfrac{2}{598}\)

=\(\dfrac{1}{3}.2\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)

=\(\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}\right)\)

=\(\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)\)

=\(\dfrac{2}{3}.\dfrac{6}{13}\)

=\(\dfrac{4}{13}\)