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a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
a)
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)
c)
\(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)
\(A=\left(4+\frac{1}{5}\right).\frac{19}{8}+\left(2+\frac{5}{8}\right).\frac{21}{5}\) =\(\frac{21}{5}.\frac{19}{8}+\frac{21}{8}.\frac{21}{5}\) =\(\frac{21}{5}.\left(\frac{19}{8}+\frac{21}{8}\right)\) = \(\frac{21}{5}.5\) =21 \(B=\frac{25}{2}.\left(3+\frac{2}{7}\right)-\frac{23}{7}.\left(5+\frac{1}{2}\right)\) =\(\frac{25}{2}.\frac{23}{7}-\frac{23}{7}.\frac{11}{2}\) =\(\frac{23}{7}.\left(\frac{25}{2}-\frac{11}{2}\right)\) =\(\frac{23}{7}.7=23\)
\(\begin{array}{l}\left( {\frac{{20}}{7}.\frac{{ - 4}}{{ - 5}}} \right) + \left( {\frac{{20}}{7}.\frac{3}{{ - 5}}} \right) = \frac{{20}}{7}.\left( {\frac{{ - 4}}{{ - 5}} + \frac{3}{{ - 5}}} \right)\\ = \frac{{20}}{7}.\left( {\frac{{ - 1}}{{ - 5}}} \right) = \frac{{20}}{7}.\frac{1}{5} = \frac{{20}}{{35}} = \frac{4}{7}\end{array}\)
\(A=\left(\frac{2}{5}+\frac{-3}{7}+\frac{1}{3}\right)+\left(-\frac{4}{7}+\frac{3}{5}\right)\)
\(=\frac{2}{5}-\frac{3}{7}+\frac{1}{3}-\frac{4}{7}+\frac{3}{5}\)
\(=\left(\frac{2}{5}+\frac{3}{5}\right)-\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{1}{3}\)
\(=1-1+\frac{1}{3}\)
\(=\frac{1}{3}\)
Ta có: \(A=\left(\frac{2}{5}+\frac{-3}{7}+\frac{1}{3}\right)+\left(\frac{-4}{7}+\frac{3}{5}\right)\)
\(\Leftrightarrow A=\frac{2}{5}+\frac{-3}{7}+\frac{1}{3}+\frac{-4}{7}+\frac{3}{5}\)
\(\Leftrightarrow A=\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{-3}{7}+\frac{-4}{7}\right)+\frac{1}{3}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{1}{3}\)
\(\Leftrightarrow A=\frac{1}{3}\)
Vậy \(A=\frac{1}{3}\)