\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)

\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)< 3.\frac{1}{3}=1\)

=> A < 1

Ta có :

\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)

\(A=3.\left(\frac{2}{3.5}\right)+3.\left(\frac{3}{5.8}\right)+3.\left(\frac{4}{8.12}\right)+3.\left(\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)\)

\(A=3.\frac{14}{51}\)

\(A=\frac{14}{17}< 1\)

Vậy A < 1

_Chúc bạn học tốt_

E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)

\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)

\(=6\cdot\frac{99}{100}=\frac{297}{50}\)

F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)

G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

E=36/1-36/7+36/7-36/13+...+36/94-36/100

  =36-36/100=891/25

đề sai thì phải

\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)

\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)

\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)

\(A=\frac{101}{120}\)

\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)

\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-...+\dfrac{1}{47}-\dfrac{1}{57}\right)\)

\(=1-\dfrac{18}{57}=\dfrac{39}{57}=\dfrac{13}{19}\)