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\(-1,53:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(=-1,53:\left(5\dfrac{5}{28}-\dfrac{17}{9}.\dfrac{5}{4}+1\dfrac{16}{63}\right)\)
\(=-1,53:\left(5\dfrac{5}{28}-\dfrac{17.5}{9.4}+1\dfrac{16}{63}\right)\)
\(=-1,53:\left(5\dfrac{5}{28}-\dfrac{85}{36}+1\dfrac{16}{63}\right)\)
\(=-1,53:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)
\(=-1,53:\left(\dfrac{1305}{252}-\dfrac{595}{252}+\dfrac{316}{252}\right)\)
\(=-1,53:\dfrac{1305-595+316}{252}\)
\(=-1,53:\dfrac{1026}{252}\)
\(=\dfrac{-153}{100}:\dfrac{57}{14}\)
\(=\dfrac{-153}{100}.\dfrac{14}{57}\)
\(=\dfrac{-153.14}{100.57}\)
\(=\dfrac{-2142}{5700}\)
\(=\dfrac{-357}{950}\)
\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(=-5,13:\left(5\dfrac{5}{28}-2\dfrac{13}{36}+1\dfrac{16}{63}\right)\)
\(=5,13:\left[\left(5-2+1\right)+\left(\dfrac{5}{28}-\dfrac{13}{36}+\dfrac{16}{63}\right)\right]\)
\(=5,13:\left(4+\dfrac{1}{14}\right)\)
\(=1,26\)
\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{125}{100}+\dfrac{79}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{5}{4}+\dfrac{79}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)
\(=-5,13:\dfrac{57}{14}=-5,13:\dfrac{15}{57}\)
\(=\dfrac{-71,82}{57}=1,26\)
Vậy \(A=1,26\)
\(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)
\(=\left(\dfrac{10}{3}.1,9+19,5:\dfrac{13}{3}\right).\left(\dfrac{62-12}{75}\right)\)
\(=\left(\dfrac{19}{3}+\dfrac{58,5}{13}\right).\dfrac{50}{75}\)
\(=\left(\dfrac{19}{3}+4,5\right).\dfrac{2}{3}\)
\(=\dfrac{32,5}{3}.\dfrac{2}{3}=\dfrac{65}{9}=7\dfrac{2}{9}\)
Vậy \(B=7\dfrac{2}{9}\)
A= -5,13: (\(5\dfrac{5}{28}\)-\(1\dfrac{8}{9}\).1,25+\(1\dfrac{16}{63}\))
A=-5,13:(\(\dfrac{145}{28}\)-\(\dfrac{17}{9}\). 1,25+\(\dfrac{79}{63}\))
A= -5,13:(\(\dfrac{145}{28}\)-\(\dfrac{85}{36}\)+ \(\dfrac{79}{63}\))
A= -5,13:\(\dfrac{57}{14}\)
A= -\(\dfrac{63}{50}\)
a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)
b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)
c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)
\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)
\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)
\(-1,53:\left(5\dfrac{5}{28}-1\dfrac{8}{9}\cdot1.25+1\dfrac{16}{63}\right)\)
\(=-\dfrac{153}{100}:\left(\dfrac{145}{28}-\dfrac{17}{9}\cdot\dfrac{5}{4}+\dfrac{79}{63}\right)\)
\(=-\dfrac{153}{100}:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)
\(=-\dfrac{153}{100}:\dfrac{1395-595+316}{252}\)
\(=-\dfrac{153}{100}\cdot\dfrac{252}{1116}=\dfrac{-153}{100}\cdot\dfrac{7}{31}=\dfrac{-1071}{3100}\)