Ta có:

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

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

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

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

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

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

A=x

B=y

h nha

A=\(\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{31.37}\)= \(\frac{1}{6}.\)(\(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{31.37}\))=\(\frac{1}{6}.\)(\(\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{31}-\frac{1}{37}\)) = \(\frac{1}{6}.\left(\frac{1}{1}-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)

ĐS: A=6/37

em làm như sau nhé : :))) 

6A= 6/7 + 6/91+...+ 6/1147

<=>6A= 6/7+1/7-1/13+1/13-1/19+...+1/31-1/37

<=> 6A= 6/7+1/7 -1/37

<=> A=6/37

Ta có : \(A=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+...+\frac{1}{1147}\)

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

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

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

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

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

Vậy \(A=\frac{6}{37}\)

chắc bạn viết sai đầu bài rồi

\(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+.......+\frac{1}{31.37}=\frac{1-\frac{1}{37}}{6}\)

=1/1×7+1/7×13+1/13×19+...+1/31×37

=1/6×(1-1/7)+1/6×(1/7-1/13)+1/6×(1/13-1/19)+...+1/6×(1/31-1/37)

=1/6×(1-1/7+1/7-1/13+1/13-1/19+...+1/31-1/37)

=1/6×(1-1/37)

=1/6×36/37

=6/37