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\(\dfrac{11}{6}+\dfrac{1}{4}=\dfrac{22}{12}+\dfrac{3}{12}=\dfrac{25}{12}\)
\(\dfrac{2}{5}-\dfrac{3}{8}=\dfrac{16}{40}-\dfrac{15}{40}=\dfrac{1}{40}\)
\(\dfrac{3}{10}-\dfrac{4}{15}=\dfrac{9}{30}-\dfrac{8}{30}=\dfrac{1}{30}\)
\(3+\dfrac{2}{5}=\dfrac{15}{5}+\dfrac{2}{5}=\dfrac{17}{5}\)
\(\dfrac{333}{777}+\dfrac{22}{55}=\dfrac{3}{7}+\dfrac{2}{5}=\dfrac{15}{35}+\dfrac{14}{35}=\dfrac{29}{35}\)
\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)
\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)
1: \(=\dfrac{7}{12}\left(-\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{7}{12}\cdot\dfrac{26}{8}+\dfrac{1}{2}=\dfrac{115}{48}\)
2: \(=\dfrac{8-3-10}{12}\cdot4+\dfrac{3}{2}\cdot\dfrac{-4}{15}\)
\(=\dfrac{-5}{3}+\dfrac{-12}{30}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-25-6}{15}=-\dfrac{31}{15}\)
3: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)
Lời giải:
a.
$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$
b.
$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$
c.
$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$
$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$
d.
$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$
$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$
e.
$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$
$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$
a) 4 + ( -1 ) 3 = 4 + ( -1 ) = 3
b ) \(\left(-\frac{1}{4}+\frac{5}{8}\right)+\left(-\frac{3}{8}\right)=\left(-\frac{2}{8}+\frac{5}{8}\right)-\frac{3}{8}=\frac{3}{8}-\frac{3}{8}=0\)
c) { [ ( -3 )2 . 10 - 2.5 - 100 ] + | -10 |: ( -2 ) } : ( -5 )
={ [ ( -9 . 10 - 10 - 100 ] + 10: ( -2 ) } : ( -5 )
={ [ -90 - 10 - 100 ] + (-5) } : ( -5 )
={ [ -100 - 100 ] + ( -5 ) } : ( -5 )
={ -200 + ( -5 ) } : ( -5 )
= -205 : ( -5 )
= 41