1.
\(-\dfrac{4}{3}+\dfrac{1}{3}:\left(-1,5\right)=-\dfrac{4}{3}+\dfrac{1}{3}:\left(-\dfrac{3}{2}\right)\)
\(=-\dfrac{4}{3}-\dfrac{2}{9}=\dfrac{-8}{9}-\dfrac{2}{9}=-\dfrac{10}{9}\)
2.
\(-\dfrac{5}{3}+\dfrac{2}{3}:\left(-1,5\right)=-\dfrac{5}{3}+\dfrac{2}{3}:\left(-\dfrac{3}{2}\right)\)
\(=-\dfrac{5}{3}+\left(-\dfrac{4}{9}\right)=-\dfrac{15}{9}+\left(-\dfrac{4}{9}\right)=-\dfrac{19}{9}\)
3.
\(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|\dfrac{-2}{3}\right|=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{3}{4}\)
\(=\dfrac{8}{12}-\dfrac{9}{12}=-\dfrac{1}{12}\)
4.
\(\left(40\%-1\dfrac{2}{3}\right):\dfrac{-3}{2}+\left(\dfrac{-1}{2}\right)^2=\left(\dfrac{2}{5}-\dfrac{5}{3}\right):\dfrac{-3}{2}+\dfrac{1}{4}\)
\(=\left(\dfrac{-19}{15}\right).\left(-\dfrac{2}{3}\right)+\dfrac{1}{4}=\dfrac{38}{45}+\dfrac{1}{4}=\dfrac{197}{180}\)
5.
\(\dfrac{5}{14}-3,7-\dfrac{19}{14}+\dfrac{8}{9}-6,3\)
\(=\left(\dfrac{5}{14}-\dfrac{19}{14}\right)-\left(3,7+6,3\right)+\dfrac{8}{9}\)
\(=\dfrac{-14}{14}-10+\dfrac{8}{9}=-1-10+\dfrac{8}{9}\)
\(=-11+\dfrac{8}{9}=\dfrac{-99}{9}+\dfrac{8}{9}=\dfrac{-91}{9}\)
6.
\(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)
\(=-3-\dfrac{16}{23}-\dfrac{2}{7}-\dfrac{7}{23}+\dfrac{9}{7}\)
\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)+\left(-\dfrac{2}{7}+\dfrac{9}{7}\right)\)
\(=-3-\dfrac{23}{23}+\dfrac{7}{7}=-3-1+1=-3\)
7.
\(-\left(\dfrac{8}{9}+\dfrac{7}{5}-\dfrac{2}{11}\right)+\left(-\dfrac{2}{5}+\dfrac{17}{9}-\dfrac{13}{11}\right)\)
\(=-\dfrac{8}{9}-\dfrac{7}{5}+\dfrac{2}{11}-\dfrac{2}{5}+\dfrac{17}{9}-\dfrac{13}{11}\)
\(=\left(-\dfrac{8}{9}+\dfrac{17}{9}\right)+\left(\dfrac{2}{11}-\dfrac{13}{11}\right)-\left(\dfrac{7}{5}+\dfrac{2}{5}\right)\)
\(=\dfrac{-9}{9}+\dfrac{-11}{11}-\dfrac{9}{5}=-1-1-\dfrac{9}{5}\)
\(=-2-\dfrac{9}{5}=-\dfrac{19}{5}\)
8.
\(\dfrac{1}{7}.\dfrac{3}{8}+\dfrac{1}{7}.\dfrac{5}{8}+\dfrac{\left(-1\right)^{2019}}{7}\)
\(=\dfrac{1}{7}.\dfrac{3}{8}+\dfrac{1}{7}.\dfrac{5}{8}+\dfrac{-1}{7}\)
\(=\dfrac{1}{7}.\left(\dfrac{3}{8}+\dfrac{5}{8}-1\right)\)
\(=\dfrac{1}{7}\left(\dfrac{8}{8}-1\right)=\dfrac{1}{7}.0\)
\(=0\)
9.
\(\dfrac{4}{5}.\dfrac{3}{7}-\dfrac{-4}{7}:\sqrt{\dfrac{25}{16}}-\left|-1\right|\)
\(=\dfrac{4}{5}.\dfrac{3}{7}+\dfrac{4}{7}:\dfrac{5}{4}-1\)
\(=\dfrac{4}{5}.\dfrac{3}{7}+\dfrac{4}{7}.\dfrac{4}{5}-1\)
\(=\dfrac{4}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-1\)
\(=\dfrac{4}{5}.1-1=-\dfrac{1}{5}\)
10.
\(\left|1,37\right|.\dfrac{2}{9}-7,37.\left|-\dfrac{2}{9}\right|\)
\(=1,37.\dfrac{2}{9}-7,37.\dfrac{2}{9}\)
\(=\dfrac{2}{9}.\left(1,37-7,37\right)\)
\(=\dfrac{2}{9}.\left(-6\right)=-\dfrac{4}{3}\)
11.
\(\left(\dfrac{-2}{3}\right)^2.\dfrac{5}{13}-\dfrac{8}{13}:\dfrac{-9}{4}-\dfrac{1}{2}\)
\(=\dfrac{4}{9}.\dfrac{5}{13}+\dfrac{8}{13}.\dfrac{4}{9}-\dfrac{1}{2}\)
\(=\dfrac{4}{9}.\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-\dfrac{1}{2}\)
\(=\dfrac{4}{9}.1-\dfrac{1}{2}=\dfrac{4}{9}-\dfrac{1}{2}\)
\(=-\dfrac{1}{18}\)
12.
\(\left(\dfrac{4}{9}\right)^{11}.\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{8}{27}\right)^{10}\)
\(=\left[\left(\dfrac{2}{3}\right)^2\right]^{11}.\left(\dfrac{2}{3}\right)^{12}:\left[\left(\dfrac{2}{3}\right)^3\right]^{10}\)
\(=\left(\dfrac{2}{3}\right)^{22}.\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{30}\)
\(=\left(\dfrac{2}{3}\right)^{34}:\left(\dfrac{2}{3}\right)^{30}\)
\(=\left(\dfrac{2}{3}\right)^4=\dfrac{16}{81}\)
13.
\(\left(\dfrac{1}{4}\right)^5.\left(\dfrac{1}{8}\right)^6:\left(\dfrac{1}{2}\right)^{26}\)
\(=\left[\left(\dfrac{1}{2}\right)^2\right]^5.\left[\left(\dfrac{1}{2}\right)^3\right]^6:\left(\dfrac{1}{2}\right)^{26}\)
\(=\left(\dfrac{1}{2}\right)^{10}.\left(\dfrac{1}{2}\right)^{18}:\left(\dfrac{1}{2}\right)^{26}\)
\(=\left(\dfrac{1}{2}\right)^{28}:\left(\dfrac{1}{2}\right)^{26}\)
\(=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
14.
\(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}=\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}.\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)
\(=\dfrac{2}{\dfrac{3}{2}}=\dfrac{4}{3}\)
