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Ta có:
\(A=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(6A=\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\)
\(=1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\)
\(=1-\frac{1}{37}=\frac{36}{37}\)
\(A=\frac{6}{37}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(A=\frac{6}{37}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+.......+\frac{1}{31.37}=\frac{1-\frac{1}{37}}{6}\)
=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
\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\times7}+\dfrac{1}{7\times13}+\dfrac{1}{13\times19}+\dfrac{1}{19\times25}+\dfrac{1}{25\times31}\)
\(=\dfrac{1}{6}\text{x}\left(\dfrac{6}{1\text{x}7}+\dfrac{6}{7\text{x}13}+...+\dfrac{6}{25\text{x}31}\right)\)
\(=\dfrac{1}{6}\text{x}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{31}\right)\)
\(=\dfrac{1}{6}\text{x}\left(1-\dfrac{1}{31}\right)=\dfrac{1}{6}\text{x}\dfrac{30}{31}=\dfrac{5}{31}\)