Tính tổng : \(1-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{3}-\dfrac{1}{28}-\dfrac{1}{6}-\dfrac{1}{21}\)
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17 tháng 7 2023
\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)
\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)
Vậy D=1/4
27 tháng 9 2021
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: \(A=\dfrac{7}{9}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 3 2023
\(F=\dfrac{5}{6}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}\left(\dfrac{225}{20}-\dfrac{37}{4}\right):\dfrac{25}{3}\)
\(F=\dfrac{5}{6}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}.\dfrac{3}{25}\)
\(F=\dfrac{5}{6}+\dfrac{41}{25}\)
\(F=\dfrac{371}{150}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 3 2023
\(D=\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\times\dfrac{21}{24}\)
\(D=\left(\dfrac{272}{30}-\dfrac{168}{30}+\dfrac{186}{30}\right)\times\dfrac{21}{24}\)
\(D=\dfrac{290}{30}\times\dfrac{21}{24}\)
\(D=\dfrac{29}{3}\times\dfrac{7}{8}\)
\(D=\dfrac{203}{24}\)
VT
25 tháng 6 2023
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
AN
10 tháng 4 2017
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=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)
GL
24 tháng 8 2023
a) \(\dfrac{4}{24}+\dfrac{7}{6}=\dfrac{4}{24}+\dfrac{28}{24}=\dfrac{4+28}{24}=\dfrac{32}{24}=\dfrac{4}{3}\)
b) \(\dfrac{10}{15}-\dfrac{1}{3}=\dfrac{10}{15}-\dfrac{5}{15}=\dfrac{10-5}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c) \(\dfrac{21}{28}-\dfrac{1}{4}=\dfrac{21}{28}-\dfrac{7}{28}=\dfrac{21-7}{28}=\dfrac{14}{28}=\dfrac{1}{2}\)
d) \(\dfrac{35}{40}+\dfrac{5}{8}=\dfrac{35}{40}+\dfrac{25}{40}=\dfrac{35+25}{40}=\dfrac{60}{40}=\dfrac{3}{2}\)
24 tháng 8 2023
\(a)\dfrac{4}{24}=\dfrac{1}{6} \\ \dfrac{1}{6}+\dfrac{7}{6}\\ =\dfrac{8}{6}=\dfrac{4}{3}\\ b)\dfrac{10}{15}=\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c)\dfrac{21}{28}=\dfrac{3}{4}\\ \dfrac{3}{4}-\dfrac{1}{4}\\ =\dfrac{2}{4}=\dfrac{1}{2}\\ d)\dfrac{35}{40}=\dfrac{7}{8}\\ \dfrac{7}{8}+\dfrac{5}{8}\\ =\dfrac{12}{8}=\dfrac{3}{2}\)
OY
27 tháng 7 2021
\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)
\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(M=2.\dfrac{3}{16}\)
\(M=\dfrac{3}{8}\)
Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)
\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)
\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)
Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)