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​​

190/47

Kết quả chính xác bằng bao nhiu 

\(=\frac{6}{8}x\left(\frac{8}{1x3x7}+\frac{8}{3x7x9}+...+\frac{8}{13x15x19}\right)\)

\(=\frac{6}{8}x\left(\frac{1}{1x3}-\frac{1}{3x7}+\frac{1}{3x7}-\frac{1}{7x9}+...+\frac{1}{13x15}-\frac{1}{15x19}\right)\)

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

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

= \(\frac{47}{190}\)