\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

\(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}.\dfrac{36}{37}\)

\(=\dfrac{6}{37}\)

\(#Wendy.Dang\)

\(\frac{-1}{91}+\frac{-1}{247}+\frac{-1}{475}+\frac{-1}{775}+\frac{-1}{1147}\)

\(=-\left(\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\right)\)

\(=-[\frac{1}{6}.\left(\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)]\)

\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\text{]}\)

\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{37}\right)\text{]}\)

\(=-\text{[}\frac{1}{6}.\frac{30}{259}\text{]}\)

\(=-\frac{5}{259}\)

Câu 2:

\(B=\frac17+\frac{1}{91}+\cdots+\frac{1}{1147}\)

\(=\frac16\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\cdots+\frac{6}{31\cdot37}\right)\)

\(=\frac16\left(1-\frac17+\frac17-\frac{1}{13}+\cdots+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac16\left(1-\frac{1}{37}\right)=\frac16\cdot\frac{36}{37}=\frac{6}{37}\)

Câu 4:

\(D=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{18\cdot19\cdot20}\)

\(=\frac12\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{18\cdot19\cdot20}\right)\)

\(=\frac12\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\cdots+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)

\(=\frac12\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)=\frac12\cdot\left(\frac12-\frac{1}{380}\right)=\frac12\cdot\frac{179}{380}=\frac{179}{760}\)

\(\dfrac{1}{7}+\dfrac{1}{91}+...+\dfrac{1}{1147}\)

\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+...+\dfrac{1}{31\cdot37}\)

\(=\dfrac{1}{6}\cdot\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{31\cdot37}\right)\)

\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{36}{37}\)

\(=\dfrac{6}{37}\)

Vậy ...

#\(Toru\)

thế này à:

\(\frac{91-\frac{1}{11}-\frac{2}{12}-\frac{3}{13}-...-\frac{91}{101}}{\frac{1}{55}+\frac{1}{60}+....+\frac{1}{505}}\)

\(\frac{91-\frac{1}{11}-\frac{2}{12}-\frac{3}{13}-...-\frac{91}{101}}{\frac{1}{55}+\frac{1}{60}+\frac{1}{65}+...+\frac{1}{505}}\)

Xét tử:

\(91-\frac{1}{11}-\frac{2}{12}-\frac{3}{13}-...-\frac{91}{101}\)

= \(\left(1+1+1+...+1\right)-\left(\frac{1}{11}+\frac{2}{12}+\frac{3}{13}+...+\frac{91}{101}\right)\)

= \(\left(1-\frac{1}{11}\right)+\left(1-\frac{2}{12}\right)+....+\left(1-\frac{91}{101}\right)\)

= \(\frac{10}{11}+\frac{10}{12}+...+\frac{10}{101}\)

= \(10.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{101}\right)\)

= \(10.5.\left(\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}\right)\)

= \(50.\left(\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}\right)\)

Thay vào ta được phân số:

\(\frac{50.\left(\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}\right)}{\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}}\)

= 50