A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)

A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)

A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)

tớ làm hơi gọn nên có gì kho hiểu thì nói tớ

=2(1/12+1/30+...+1/132)

=2(1/3-1/4+1/5-1/6+1/6-1/7+...+1/11-1/12)

=2(1/12+1/5-1/12)

=2*1/5=2/5

Ta có:

\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)

\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)

\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)

\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)

\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)

\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)

\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)

    \(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)

    \(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)

    \(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)

    \(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)

    \(=\dfrac{28}{36}=\dfrac{7}{9}\)

Vậy: \(A=\dfrac{7}{9}\)

\(A=4,8.\left(3,1-1,5\right)+1,5.\left(4,8-3,1\right)\)

\(A=4,8.3,1-4,8.1,5+1,5.4,8-1,5.3,1\)

\(A=3,1.\left(4,8-1,5\right)-4,8\left(1,5+1,5\right)\)

\(A=3,1.3,3-4,8.3\)

\(A=10,23-14,4=-4,17\)

\(B=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2.3^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+3^9.2^{18}.5}{2^{11}.3^9+3^{10}.2^{20}}=\dfrac{2^{18}.3^9\left(2+5\right)}{2^{11}.3^9\left(1+3.2^9\right)}=\dfrac{2^7.7}{1+3.2^9}\)

\(C=-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}-\dfrac{1}{36}-\dfrac{1}{45}\)

\(\Rightarrow C=\dfrac{-1}{2}\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}\left(1-\dfrac{1}{10}\right)\)

\(\Rightarrow C=\dfrac{-1}{2}.\dfrac{9}{10}\)

\(\Rightarrow C=\dfrac{-9}{20}\)

\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)

\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)

\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)

Vậy D=1/4

\(\dfrac{1}{2}N=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)

N=\(\dfrac{2}{5}:\dfrac{1}{2}=\dfrac{4}{5}\)

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)

\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)

\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)

Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)

\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)

\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(M=2.\dfrac{3}{16}\)

\(M=\dfrac{3}{8}\)

Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)