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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: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(=\frac{\frac{1}{7}\left(\frac{1}{2}-\sqrt{2}+\frac{3\sqrt{2}}{5}\right).\frac{-4}{15}}{\frac{1}{5}\left(\frac{1}{2}+\frac{3\sqrt{2}}{5}-\sqrt{2}\right).\frac{5}{7}}\)
\(=\frac{\frac{1}{7}.\frac{-4}{15}}{\frac{1}{5}.\frac{5}{7}}=\frac{\frac{-4}{105}}{\frac{1}{7}}=\frac{-4}{15}\)
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3.\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{10}.\frac{5}{7}+\frac{3.\sqrt{2}}{25}.\frac{5}{7}-\frac{\sqrt{2}}{5}.\frac{5}{7}}\)
\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{14}+\frac{3.\sqrt{2}}{35}-\frac{\sqrt{2}}{7}}\)
\(=\frac{-4}{15}\)
Học tốt
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
