Ta có:

\(A=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

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

\(6A=\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\)

\(=1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\)

\(=1-\frac{1}{37}=\frac{36}{37}\)

\(A=\frac{6}{37}\)

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

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

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

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

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

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

Tổng cần tính bằng:\(\frac{1}{1.7}\)+\(\frac{1}{7.13}\)+\(\frac{1}{13.19}\)+\(\frac{1}{19.25}\)+\(\frac{1}{25.31}\)+\(\frac{1}{31.37}\)=(\(\frac{1}{1}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{13}\)+...+\(\frac{1}{31}\)\(\frac{1}{37}\)):3 =(\(1\)-\(\frac{1}{37}\)):3=\(\frac{12}{37}\)

\(\frac{12}{37}\)kà kết quả nha bạn. chúc bạn học tốt. ( kik mình nha)

A=1/7 +1/91 +1/247 + 1/475 + 1/775 + 1/1147

A=1/(1.7)+1/(7.13)+1/(13.19)+...+1/(31...  

A=(1/6)*( 1 - 1/7 + 1/7 - 1/13 +... +1/31-1/37)  

A=(1/6)*(1-1/37)  

A=(1/6)*(36/37)  

A=6/37 

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

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

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

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

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

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

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

C=\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

C=\(\frac{1}{6}\left\{\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{31.37}\right\}\)=\(\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}+....+\frac{1}{31}-\frac{1}{37}\right)\)

C=\(\frac{1}{6}\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{36}{222}=\frac{6}{37}\)

D=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{49.51}\)

D=\(\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

D=\(\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

D=\(\frac{3}{2}\left(1-\frac{1}{51}\right)=\frac{3}{2}.\frac{50}{51}\)

D=\(\frac{150}{102}\)=\(\frac{25}{17}\)

= \(\frac{6}{37}\)