$a)-12,4-2,4:\left(-2\frac{2}{3}\right)+2\frac{2}{3}\cdot \left(-0,9\right)-0,1=$

$=-12,4-\frac{24}{10}:\left(-\frac{8}{3}\right)+\frac{8}{{\sout{3}}^1}\cdot \left(-\frac{{\sout{9}}^3}{10}\right)-0,1=$

$=-12,4-\frac{{\sout{12}}^3}{5}\cdot \left(-\frac{3}{{\sout{8}}^2}\right)+\left(-\frac{24}{10}\right)-0,1=$

$=-12,4-\left(-\frac{9}{10}\right)-2,4-0,1=$

$=-12,4+0,9-2,4-0,1=$

$=-11,5-2,5=-14$

 

$b)-\left[-\left(-7\frac{1}{3}\right):1,1-1,1\cdot 6\frac{2}{3}\right]+4\frac{1}{6}\cdot \left(-4\right)=$

$=-\left[7\frac{1}{3}:\frac{11}{10}-\frac{11}{{\sout{10}}^1}\cdot \frac{{\sout{20}}^2}{3}\right]+\frac{25}{{\sout{6}}^3}\cdot \left(-{\sout{4}}^2\right)=$

$=-\left[\frac{{\sout{22}}^2}{3}\cdot \frac{10}{{\sout{11}}^1}-\frac{22}{3}\right]+\left(-\frac{50}{3}\right)=$

$=-\left[\frac{20}{3}-\frac{22}{3}\right]-\frac{50}{3}=$

$=-\left[-\frac{2}{3}\right]-\frac{50}{3}=\frac{2}{3}-\frac{50}{3}=-\frac{48}{3}=-16$

 

$c)0,024\cdot \left(-3\frac{1}{3}\right):0,004-\left[-1\frac{1}{7}\cdot \left(-5\frac{5}{8}+2\frac{1}{8}:1,7\right)\right]=$

$=\frac{{\sout{24}}^8}{{\sout{1000}}^{100}}\cdot \left(-\frac{{\sout{10}}^1}{{\sout{3}}^1}\right):\frac{4}{1000}-\left[-\frac{8}{7}\cdot \left(-\frac{45}{8}+\frac{17}{8}:\frac{17}{10}\right)\right]=$

$=-\frac{{\sout{8}}^2}{{\sout{100}}^1}\cdot \frac{{\sout{1000}}^{10}}{{\sout{4}}^1}-\left[-\frac{8}{7}\cdot \left(-\frac{45}{8}+\frac{{\sout{17}}^1}{8}\cdot \frac{10}{{\sout{17}}^1}\right)\right]=$

$=-20-\left[-\frac{8}{7}\cdot \left(-\frac{45}{8}+\frac{10}{8}\right)\right]=$

$=-20-\left[-\frac{{\sout{8}}^1}{7}\cdot \left(-\frac{35}{{\sout{8}}^1}\right)\right]=$

$=-20-\frac{35}{7}=-20-5=-25$

 

$d)40-2\frac{4}{7}:\left[-0,12+\left(-\frac{3}{4}\right):12,5\right]\cdot \left(-1,4\right)=$

$=40-\frac{18}{7}:\left[-0,12+\left(-0,75\right):12,5\right]\cdot \left(-\frac{14}{10}\right)=$

$=40-\frac{18}{7}:\left[-0,12-75:1250\right]\cdot \left(-\frac{7}{5}\right)=$

$=40-\frac{18}{7}:\left[-0,12-\frac{75}{1250}\right]\cdot \left(-\frac{7}{5}\right)=$

$=40-\frac{18}{7}:\left[-0,12-\frac{3}{50}\right]\cdot \left(-\frac{7}{5}\right)=$

$=40-\frac{18}{7}:\left[-0,12-0,06\right]\cdot \left(-\frac{7}{5}\right)=$

$=40-\frac{18}{7}:\left(-0,18\right)\cdot \left(-\frac{7}{5}\right)=$

$=40-\frac{18}{7}:\left(-\frac{18}{100}\right)\cdot \left(-\frac{7}{5}\right)=$

$=40-\frac{18}{{\sout{7}}^1}\cdot \left(-\frac{100}{18}\right)\cdot \frac{-{\sout{7}}^1}{5}=$

$=40-{\sout{18}}^1\cdot \frac{100}{{\sout{18}}^1}\cdot \frac{1}{5}=$

$=40-\frac{100}{5}=40-20=20$