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\(2.B=\dfrac{2}{6}+\dfrac{2}{14}+\dfrac{2}{60}+...+\dfrac{2}{990}\)
\(2B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{9.10.11}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{9.10}-\dfrac{1}{10.11}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{10.11}\)
\(B=\dfrac{27}{110}\)
a, \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
= \(\dfrac{28}{15}.\dfrac{25}{100}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
= \(\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{32-79}{60}\right).\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-47}{60}.\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-24}{60}=\dfrac{60}{60}=1\)
b, \(\dfrac{\left(\dfrac{11^2}{200}+0,415\right):0,01}{\dfrac{1}{12}-37,25+3\dfrac{1}{6}}\)
= \(\dfrac{\left(\dfrac{121}{200}+\dfrac{415}{1000}\right):\dfrac{1}{100}}{\dfrac{1}{12}-\dfrac{3725}{100}+\dfrac{19}{6}}=\dfrac{\left(\dfrac{121}{200}+\dfrac{83}{200}\right).100}{\dfrac{1}{12}-\dfrac{149}{4}+\dfrac{19}{6}}\)
= \(\dfrac{\dfrac{51}{50}.100}{-34}=\dfrac{102}{-34}=-3\)
\(B=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)
\(\Rightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\dfrac{10}{33}\)
\(\Rightarrow B=\dfrac{10}{99}\)
Vậy...
\(B=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(\Leftrightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+..+\dfrac{1}{30.33}\)
\(\Leftrightarrow B=\left(\dfrac{1}{3}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{30}-\dfrac{1}{33}\)
\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{33}\)
\(\Leftrightarrow B=\dfrac{10}{33}\).
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
\(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}\)
\(A=\frac{19}{24}+\frac{-1}{2}+\frac{-1}{3}+\frac{-7}{24}\)
\(A=\left(\frac{19}{24}+\frac{-7}{24}\right)+\frac{-1}{2}+\frac{-1}{3}\)
\(A=\frac{1}{2}+\frac{-1}{2}+\frac{-1}{3}\)
\(A=0+\frac{-1}{3}=\frac{-1}{3}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{15}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}+\frac{-3}{7}+\frac{-1}{3}\)
\(B=\left(\frac{49}{168}+\frac{140}{168}+\frac{42}{168}\right)+\left(\frac{-72}{168}+\frac{-56}{168}\right)\)
\(B=\frac{231}{168}+\frac{-128}{168}=\frac{103}{168}\)
Có: \(A=\frac{-1}{3}=\frac{\left(-1\right)\cdot56}{3\cdot56}=\frac{-56}{168}\)
Mặt khác: \(-56< 103\)
\(\Rightarrow\)\(\frac{-56}{168}< \frac{103}{168}\)
\(hay\) \(A< B\)
Ta có:\(B=\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+...+\dfrac{1}{990}\)
\(2B=\dfrac{2}{6}+\dfrac{2}{24}+\dfrac{2}{60}+...+\dfrac{2}{990}\)
\(2B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{9\cdot10\cdot11}\)
\(2B=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+...+\dfrac{1}{9\cdot10}-\dfrac{1}{10\cdot11}\)
\(2B=\dfrac{1}{1\cdot2}-\dfrac{1}{10\cdot11}\)
\(2B=\dfrac{27}{55}\)
\(B=\dfrac{27}{55}:2\)
\(B=\dfrac{27}{110}\)