Giải:

Đặt:

\(A=\dfrac{6}{1.3.7}+\dfrac{6}{3.7.9}+\dfrac{6}{7.9.13}+\dfrac{6}{9.13.15}+\dfrac{6}{13.15.19}\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{8}{1.3.7}+\dfrac{8}{3.7.9}+\dfrac{8}{7.9.13}+\dfrac{8}{9.13.15}+\dfrac{8}{13.15.19}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{1.3}-\dfrac{1}{3.7}+\dfrac{1}{3.7}-\dfrac{1}{7.9}+\dfrac{1}{7.9}-\dfrac{1}{9.13}+\dfrac{1}{9.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.19}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{1.3}-\dfrac{1}{15.19}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{3}-\dfrac{1}{285}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}.\dfrac{94}{285}\)

\(\Leftrightarrow A=\dfrac{47}{190}\)

Vậy ...

B=6/1.3.5+6/3.7.9+6/7.9.13+6/9.13.15+6/13.15.19

 =2/5+2/63+2/273+2/585+2/1235

 =886/1995

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19

\(=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)\)

\(=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)

\(=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)\)

\(=\frac{6}{8}\cdot\frac{94}{285}\)

\(=\frac{47}{190}\)

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=​8​​6​​(​1.3.7​​8​​+​3.7.9​​8​​+...+​13.15.19​​8​​)
=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=​8​​6​​(​1.3​​1​​−​3.7​​1​​+​3.7​​1​​−​7.9​​1​​+...+​13.15​​1​​−​15.19​​1​​)
=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=​8​​6​​⋅(​3​​1​​−​285​​1​​)
=\frac{6}{8}\cdot\frac{94}{285}=​8​​6​​⋅​285​​94​​
=\frac{47}{190}=​190​​47​​

kq 47/190

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=68(81.3.7+83.7.9+...+813.15.19)=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=​8​​6​​(​1.3.7​​8​​+​3.7.9​​8​​+...+​13.15.19​​8​​)
=68(11.3−13.7+13.7−17.9+...+113.15−115.19)=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=​8​​6​​(​1.3​​1​​−​3.7​​1​​+​3.7​​1​​−​7.9​​1​​+...+​13.15​​1​​−​15.19​​1​​)
=68⋅(13−1285)=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=​8​​6​​⋅(​3​​1​​−​285​​1​​)
=68⋅94285=\frac{6}{8}\cdot\frac{94}{285}=​8​​6​​⋅​285​​94​​
=47190=\frac{47}{190}=​190​​47​​

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=​8​​6​​(​1.3.7​​8​​+​3.7.9​​8​​+...+​13.15.19​​8​​)
=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=​8​​6​​(​1.3​​1​​−​3.7​​1​​+​3.7​​1​​−​7.9​​1​​+...+​13.15​​1​​−​15.19​​1​​)
=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=​8​​6​​⋅(​3​​1​​−​285​​1​​)
=\frac{6}{8}\cdot\frac{94}{285}=​8​​6​​⋅​285​​94​​
=\frac{47}{190}=​190​​47​​

c=94/285

mình cũng ko hiểu cách giải này cho lắm nên mình làm thế thôi

X=1

TK MINH NHA

\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)

\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=2.\frac{15}{32}\)

\(S=\frac{15}{16}< 1\RightarrowĐPCM\)

Vậy \(S=\frac{15}{16}\)