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Câu C giải rồi
\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)
\(\frac{1}{5}+\frac{1}{20}+\frac{1}{44}+...+\frac{1}{170}=\frac{2}{10}+\frac{2}{40}+\frac{2}{88}+...+\frac{2}{340}\)
\(\frac{2}{2x5}+\frac{2}{5x8}+\frac{2}{8x11}+...+\frac{2}{17x20}=2x\left(\frac{1}{2x5}+\frac{1}{5x8}+\frac{1}{8x11}+...+\frac{1}{17x20}\right)\)
\(2x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)=2x\(\left(\frac{1}{2}-\frac{1}{20}\right)\)
2x\(\frac{9}{20}=\frac{9}{10}\)
S=1/20+1/44+1/77+...+1/3080
S=1/20+1/44+1/77+...+1/3080
S.3/2=3/40+3/88+3/154+...+3/6160
S.3/2=3/40+3/88+3/154+...+3/6160
S.3/2=3/5.8+3/8.11+3/11.14+...+3/77.80
S.3/2=3/5.8+3/8.11+3/11.14+...+3/77.80
S.3/2=1/5−1/8+1/8−1/11+1/11−1/14+...+1/77−1/80
S.3/2=1/5-1/8+1/8-1/11+1/11-1/14+...+1/77-1/80
S.3/2=1/5−1/80
S.3/2=1/5-1/80
S.3/2=3/16
S.3/2=3/16
S=3/16:3/2
S=3/16:3/2
S=1/8
\(\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
=\(\dfrac{2}{10}+\dfrac{2}{40}+\dfrac{2}{88}+\dfrac{2}{154}+\dfrac{2}{238}+\dfrac{2}{340}+\dfrac{2}{460}+\dfrac{2}{598}\)
=\(\dfrac{1}{3}.2\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
=\(\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
=\(\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)\)
=\(\dfrac{2}{3}.\dfrac{6}{13}\)
=\(\dfrac{4}{13}\)
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