Đặt A=\(\frac{-1}{2\cdot3\cdot4}\)+\(\frac{-1}{3\cdot4\cdot5}\)+...+\(\frac{-1}{12\cdot13\cdot14}\)

        A=(-1)*(\(\frac{1}{2\cdot3\cdot4}\)+\(\frac{1}{3\cdot4\cdot5}\)+...+\(\frac{1}{12\cdot13\cdot14}\))

        A=\(\frac{-1}{2}\)*(\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{12\cdot13\cdot14}\))

        A=\(\frac{-1}{2}\)*(\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{12\cdot13}\)-\(\frac{1}{13\cdot14}\))

       A=\(\frac{-1}{2}\)*(\(\frac{1}{6}\)-\(\frac{1}{182}\))

       A=\(\frac{-1}{2}\)*\(\frac{44}{273}\)

       A=\(\frac{-22}{273}\)

\(\frac{-1}{2.3.4}+\frac{-1}{3.4.5}+\frac{-1}{4.5.6}+...+\frac{-1}{11.12.13}+\frac{-1}{12.13.14}\)

\(=-\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{11.12.13}+\frac{2}{12.13.14}\right)\)

\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{11.12}-\frac{1}{12.13}+\frac{1}{12.13}-\frac{1}{13.14}\right)\)

\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{13.14}\right)=-\frac{1}{2}.\frac{44}{273}=-\frac{22}{273}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.50}\)

    \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

    \(=1-\frac{1}{50}\)

      \(=\frac{49}{50}\)

5,có thể bn sai đề chỗ \(\frac{1920}{111.2013}\),phải là \(\frac{1902}{111.2013}\)

\(F=\frac{3}{15.18}+\frac{11}{18.29}+\frac{12}{29.41}+\frac{70}{41.111}+\frac{1902}{111.2013}\)

\(=\frac{18-15}{15.18}+\frac{29-18}{18.29}+\frac{41-29}{12}+\frac{111-41}{41.111}+\frac{2013-111}{1902}\)

\(=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}+\frac{1}{41}-\frac{1}{111}+\frac{1}{111}-\frac{1}{2013}\)

\(=\frac{1}{15}-\frac{1}{2013}=\frac{222}{3355}\)

Ta có công thức:

\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)

vậy

\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)

\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)

\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)

mình làm đầu tiên đó, 

Chúc bạn học tốt !

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)

\(=\frac{1}{2}\cdot\frac{77}{156}\)

\(=\frac{77}{312}\)

Câu 1: 

Đặt S = 1.2+2.3+3.4+...+30.31

  3 S  = 1.2.3+2.3.3+3.4.3+...+30.31.3

  3 S  = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ...+ 30.31.(32-29)

  3S  = 1.2.3 + 2.3.4-2.3 + 3.4.5-2.3.4 + ...+ 30.31.32-29.30.31

  3S= 30.31.32 

S= 30.31.32/3

\(\frac{4}{7}x+\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{12.13.14}\right)=\frac{39}{40}\)

\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{12.13}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)

\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)

\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\frac{45}{91}\right]=\frac{39}{40}\)

\(\Leftrightarrow\frac{4}{7}x+\frac{45}{182}=\frac{39}{40}\)

\(\Leftrightarrow\frac{4}{7}x=\frac{39}{40}-\frac{45}{182}\Leftrightarrow\frac{4}{7}x=\frac{2649}{3640}\)

\(\Rightarrow x=\frac{2649}{3640}\div\frac{4}{7}=\frac{2649}{2080}\)

Vậy x = \(\frac{2649}{2080}\)