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p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)
\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)
\(=\dfrac{11}{13}\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))3
A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)
A = \(\dfrac{-73}{50}\)
B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))
B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)
B = \(\dfrac{13}{12}\)
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
a: \(=\left(\dfrac{19}{6}-\dfrac{2}{5}\right):\left(\dfrac{29}{6}+\dfrac{7}{10}\right)\)
\(=\dfrac{19\cdot5-2\cdot6}{30}:\dfrac{290+42}{30}=\dfrac{83}{332}=\dfrac{1}{4}\)
b: \(=\dfrac{\left(\dfrac{102}{25}-\dfrac{2}{25}\right)\cdot\dfrac{17}{4}}{\left(6+\dfrac{5}{9}-3-\dfrac{1}{4}\right)\cdot\dfrac{16}{7}}\)
\(=\dfrac{4\cdot\dfrac{17}{4}}{\dfrac{16}{7}\cdot\dfrac{119}{36}}=\dfrac{17}{\dfrac{68}{9}}=17\cdot\dfrac{9}{68}=\dfrac{9}{4}\)
c: \(=\left(\dfrac{120}{60}-\dfrac{15}{60}+\dfrac{20}{60}-\dfrac{36}{60}\right):\left(\dfrac{45}{15}-\dfrac{3}{15}-\dfrac{25}{15}\right)\)
\(=\dfrac{89}{60}:\dfrac{17}{15}=\dfrac{89}{60}\cdot\dfrac{15}{17}=\dfrac{89}{68}\)
\(\left(-\dfrac{1}{3}+\dfrac{5}{6}\right)^2.4-2\)
= \(\left(\dfrac{1}{2}\right)^2.4-2\)
= \(\dfrac{1}{4}.4-2\)
= \(1-2\)
= \(-1\)
( -1/3 + 5/6 )2 . 4 - 2
= ( 1/9 + 25/36 ) . 4 -2
= 29/36 . 4 - 2
= 29/9 - 2
= 11/9