\(D=\dfrac{2}{3}.\dfrac{5}{5.8}+\dfrac{11}{8.19}+\dfrac{13}{19.32}\)

\(D=\dfrac{2.5}{2.5.8}+\dfrac{1}{19}.\dfrac{11}{8}+\dfrac{1}{19}.\dfrac{13}{32}\)

\(D=\dfrac{1}{12}+\dfrac{1}{19}.\left(\dfrac{11}{8}+\dfrac{13}{32}\right)\)

\(D=\dfrac{1}{12}+\dfrac{1}{19}.\left(\dfrac{44}{32}+\dfrac{13}{32}\right)\)

\(D=\dfrac{1}{12}+\dfrac{1}{19}.\dfrac{57}{32}\)

\(D=\dfrac{1}{12}+\dfrac{3}{32}\)

\(D=\dfrac{8}{96}+\dfrac{9}{96}\)

\(D=\dfrac{17}{96}\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )

\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)

\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)

\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)

Rồi làm tương tự cân b nha!

\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)

\(+\dfrac{1}{57}-\dfrac{1}{87}\)

\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)

\(=\dfrac{4}{1\cdot5}-\left(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{70}{31\cdot101}+\dfrac{99}{101\cdot200}\right)\)

\(=1-\dfrac{1}{5}-\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\right)\)

\(=1-\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{200}\)

\(=\dfrac{201}{200}-\dfrac{2}{5}=\dfrac{201-80}{200}=\dfrac{121}{200}\)

\(A=\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)

\(=\frac{1}{3}-\frac{1}{87}=\frac{29}{87}-\frac{1}{87}=\frac{28}{87}\)

@Minh Triều: Bạn chắc đúng ko?

\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)

\(=\frac{5-3}{3.5}+\frac{8-5}{3}+\frac{19-8}{8.19}+\frac{32-29}{19.32}+\frac{57-32}{32.57}+\frac{87-57}{57.87}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)

\(=\frac{1}{3}-\frac{1}{87}=\frac{28}{87}\)

\(C=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)\(C=\left(\dfrac{5-3}{3\cdot5}\right)+\left(\dfrac{8-5}{5\cdot8}\right)+\left(\dfrac{19-8}{8\cdot19}\right)+\left(\dfrac{32-19}{19\cdot32}\right)+\left(\dfrac{57-32}{32\cdot57}\right)+\left(\dfrac{87-57}{57\cdot87}\right)\)\(C=\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{19}\right)+\left(\dfrac{1}{19}-\dfrac{1}{32}\right)+\left(\dfrac{1}{32}-\dfrac{1}{57}\right)+\left(\dfrac{1}{57}+\dfrac{1}{87}\right)\)\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\)\(C=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)

\(S=\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.85}\)

\(S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)

\(S=\frac{1}{3}-\frac{1}{87}\)

\(S=\frac{28}{87}\)

cậu lấy máy tính,tính đi ạ

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