\(D=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+\frac{4}{285}+\frac{4}{437}+\frac{4}{621}\)
K
Khách
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K
2
9 tháng 6 2017
A = \(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+\frac{1}{285}+\frac{1}{437}\)
A = \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}\)
A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}.\left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{4}.\left(\frac{1}{11}-\frac{1}{15}\right)+\frac{1}{4}.\left(\frac{1}{15}-\frac{1}{19}\right)+\frac{1}{4}.\left(\frac{1}{19}-\frac{1}{23}\right)\)
A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}\right)\)
A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{23}\right)\)
A = \(\frac{1}{4}.\frac{20}{69}\)
A = \(\frac{5}{69}\)
U
24 tháng 7 2018
\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{28.\left(\frac{5}{84}+\frac{3}{285}+\frac{4}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{1}{77}+\frac{1}{165}+\frac{1}{285}\right)}{25.4.\left(\frac{5}{4.84}+\frac{3}{4.180}+\frac{4}{4.285}\right)}\)
\(x-\frac{27}{100}=\frac{73,\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{5}{366}+\frac{3}{240}+\frac{4}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{2}{105}+\frac{1}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73}{100}\)
\(x=\frac{27}{100}+\frac{73}{100}\)
\(\Rightarrow x=1\)
1 tháng 5 2017
\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{25.\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{1}{77}+\frac{1}{165}+\frac{1}{285}\right)}{25.4.\left(\frac{5}{4.84}+\frac{3}{4.180}+\frac{4}{4.285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{5}{336}+\frac{1}{240}+\frac{1}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73.\left(\frac{2}{105}+\frac{1}{285}\right)}{100.\left(\frac{2}{105}+\frac{1}{285}\right)}\)
\(x-\frac{27}{100}=\frac{73}{100}\)
\(x=\frac{73}{100}+\frac{27}{100}\)
\(x=1\)
Vậy x=1
YY
7 tháng 7 2015
\(\left(\frac{4}{27}+\frac{4}{165}+\frac{4}{285}\right):\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)=\frac{4}{27}+\frac{4}{165}+\frac{4}{285}:\frac{5}{84}+\frac{3}{180}+\frac{4}{285}=\frac{1052}{5643}:\frac{12}{133}=\frac{1841}{891}\)
KT
3 tháng 8 2018
\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{223}+\frac{4}{437}+\frac{2}{575}\)
\(=\frac{2}{5.7}+\frac{2}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
\(D=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+\frac{4}{285}+\frac{4}{437}+\frac{4}{621}\)
\(D=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)
\(D=\frac{1}{3}-\frac{1}{27}\)
\(D=\frac{9}{27}-\frac{1}{27}\)
\(D=\frac{8}{27}\)
\(D=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+\frac{4}{285}+\frac{4}{437}+\frac{4}{621}\)
\(D=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)
\(D=\frac{1}{3}-\frac{1}{27}\)
\(D=\frac{8}{27}\)
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