a)

$\left(\frac{3}{4}+\frac{2}{3}\right)\cdot 6=\left(\frac{9}{12}+\frac{8}{12}\right)\cdot 6=\frac{17}{{\sout{12}}^2}\cdot {\sout{6}}^1=\frac{17}{2}=8\frac{1}{2}$

 

b)

$1\frac{5}{6}-{\sout{2}}^1\cdot \frac{3}{{\sout{4}}^2}=1\frac{10}{12}-\frac{3}{2}=1\frac{10}{12}-1\frac{6}{12}=\frac{4}{12}=\frac{1}{3}$

 

c)

$1\frac{7}{10}\cdot 10+3\cdot 1\frac{1}{12}=\frac{17}{{\sout{10}}^1}\cdot {\sout{10}}^1+3\frac{3}{12}=17+3\frac{1}{4}=20\frac{1}{4}$