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M = (345 x (6789 + 3456 - 245)/690) x 99/100 x 98/99 x...x 2/3 x 1/2
M = ((345 x 10000)/690) x 99/2 (rút gọn)
M = (10000/2) x 99/2
M = 5000 x 99/2
M = 247500
Ok nha
a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{11}{10}+\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+5\times\dfrac{5}{8}=\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\times\dfrac{25}{29}=\left(\dfrac{14}{35}+\dfrac{15}{35}\right)\times\dfrac{25}{39}=\dfrac{29}{35}\times\dfrac{25}{39}=\dfrac{145}{274}\)
d) \(\dfrac{1}{4}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{4}{5}=\dfrac{5}{12}\times\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}\times\dfrac{21}{20}=\dfrac{105}{240}=\dfrac{7}{16}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}x\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\dfrac{29}{35}x\dfrac{25}{29}=\dfrac{5}{7}\)
\(=\dfrac{5}{12}x\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}x\dfrac{21}{20}=\dfrac{7}{16}\)
a: =2/5+3/8=16/40+15/40=31/40
b: =1/2x(25/10-6/10)=1/2x19/10=19/20
a) \(\left(\dfrac{1}{4}+\dfrac{1}{12}\right):\dfrac{1}{13}\)
\(=\left(\dfrac{3}{12}+\dfrac{1}{12}\right):\dfrac{1}{13}\)
\(=\dfrac{4}{12}\times13\)
\(=\dfrac{1}{3}\times13\)
\(=\dfrac{13}{3}\)
b) \(\dfrac{3}{5}:\dfrac{2}{9}-\dfrac{1}{10}\)
\(=\dfrac{3}{5}\times\dfrac{9}{2}-\dfrac{1}{10}\)
\(=\dfrac{27}{10}-\dfrac{1}{10}\)
\(=\dfrac{26}{10}\)
\(=\dfrac{13}{5}\)
a) $\frac{3}{2} \times \frac{5}{8} + \frac{7}{4} = \frac{{15}}{{16}} + \frac{7}{4} = \frac{{15}}{{16}} + \frac{{28}}{{16}} = \frac{{43}}{{16}}$
b) $\frac{8}{5}:\left( {\frac{4}{3} - \frac{5}{6}} \right) = \frac{8}{5}:\left( {\frac{8}{6} - \frac{5}{6}} \right) = \frac{8}{5}:\frac{1}{2} = \frac{8}{5} \times 2 = \frac{{16}}{5}$
c) $\frac{3}{4} \times \frac{1}{5} - \frac{1}{{10}} = \frac{3}{{20}} - \frac{1}{{10}} = \frac{3}{{20}} - \frac{2}{{20}} = \frac{1}{{20}}$
(1 + \(\dfrac{1}{2005}\))\(x\)(1 +\(\dfrac{1}{2006}\))\(x\)(1+ \(\dfrac{1}{2007}\))(1+\(\dfrac{1}{2008}\))\(x\)(1 +\(\dfrac{1}{2009}\))
= \(\dfrac{2006}{2005}\)\(x\).\(\dfrac{2007}{2006}\)\(x\).\(\dfrac{2008}{2007}\)\(x\)\(\dfrac{2009}{2008}\)\(x\)\(\dfrac{2010}{2009}\)
= \(x\).\(x.x.x\).\(\dfrac{402}{401}\)
b; (1 - \(\dfrac{3}{4}\))\(x\)(1 - \(\dfrac{3}{7}\))\(x\)(1 - \(\dfrac{3}{10}\))\(x\)(1 - \(\dfrac{3}{13}\))\(x\)(1 - \(\dfrac{3}{16}\))
= \(\dfrac{1}{4}\)\(x\).\(\dfrac{4}{7}\)\(x\).\(\dfrac{7}{10}\)\(x\).\(\dfrac{10}{13}\)\(x\).\(\dfrac{13}{16}\)
= (\(\dfrac{1}{4}\).\(\dfrac{4}{7}\).\(\dfrac{7}{10}\).\(\dfrac{10}{13}\).\(\dfrac{13}{16}\)).(\(x.x.x.x\))
= \(\dfrac{1}{16}\).\(x.x.x.x\)
a: \(\left(1+\dfrac{1}{2005}\right)\times\left(1+\dfrac{1}{2006}\right)\times...\times\left(1+\dfrac{1}{2009}\right)\)
\(=\dfrac{2006}{2005}\times\dfrac{2007}{2005}\times...\times\dfrac{2010}{2009}\)
\(=\dfrac{2010}{2005}=\dfrac{402}{401}\)
b: \(\left(1-\dfrac{3}{4}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{13}\right)\times\left(1-\dfrac{3}{16}\right)\)
\(=\dfrac{1}{4}\times\dfrac{4}{7}\times\dfrac{7}{10}\times\dfrac{10}{13}\times\dfrac{13}{16}\)
\(=\dfrac{1}{16}\)
c: \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{8}{64}+\dfrac{4}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)
\(=\dfrac{32+16+8+4+2+1}{64}=\dfrac{63}{64}\)
d: \(2018\times\left(\dfrac{1}{2}+\dfrac{1212}{2424}\right)\)
\(=2018\times\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(=2018\times\dfrac{2}{2}=2018\)
c; C = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2 x C = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)
2xC - C = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\) - \(\dfrac{1}{64}\)
C x ( 2- 1) = (1 - \(\dfrac{1}{64}\)) + (\(\dfrac{1}{2}\)-\(\dfrac{1}{2}\)) + (\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\)) + ...+ (\(\dfrac{1}{32}\)-\(\dfrac{1}{32}\))
C = \(\dfrac{63}{64}\) + 0 + 0 + 0 + ...+ 0
C = \(\dfrac{63}{64}\)