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\(A=\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-4\dfrac{1}{5}\right)-\dfrac{11}{31}\\ =\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{31}{11}\\ =\left(\dfrac{30}{6}+\dfrac{15}{6}\right):\left(\dfrac{95}{30}-\dfrac{126}{30}\right)-\dfrac{31}{11}\\ =\dfrac{45}{6}:\dfrac{-21}{30}-\dfrac{31}{11}\\ =\dfrac{15}{2}\times\dfrac{-10}{7}-\dfrac{31}{11}=-\dfrac{75}{7}-\dfrac{31}{11}=-\dfrac{825}{77}-\dfrac{217}{77}=\dfrac{-1042}{77}\)
\(B=\left(-6\right).10:\left[-0,25+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right)+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{4}+\dfrac{-1}{4}\right)+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{2}\right)+1\dfrac{3}{4}=120+1\dfrac{3}{4}=121\dfrac{3}{4}\)
a)
\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)
b)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)
c)
\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)
d)
\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)
a)
\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\= - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)
d)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ = - \frac{3}{4}\end{array}\).
C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)
C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)
C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)
C = \(\frac{-233}{135}\)
D = \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)
D = \(-4.\frac{12}{13}\)
D = \(\frac{-48}{13}\)
E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)
E = \(5.4-4.3+5-0,3.20\)
E = \(20-12+5-6\)
E = \(8+\left(-1\right)\)
E = \(7\)
F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\)
F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)
F = \(\frac{-11}{12}\)
Chúc cậu hk tốt ~
A \(=\frac{35}{6}.\left(\frac{-30}{31}\right)-\frac{11}{31}=\frac{-175}{31}-\frac{11}{31}=\frac{-186}{31}=-6\)
\(B=-60:\left(\frac{-1}{2}\right)+\frac{7}{4}=30+\frac{7}{4}=\frac{127}{4}\)