[tex]a) \ (\dfrac{5}{6}-\dfrac{1}{3})\cdot2\dfrac{1}{4}=(\dfrac{5}{6}-\dfrac{2}{6})\cdot\dfrac{9}{4}=\dfrac{3}{6}\cdot\dfrac{9}{4}=\dfrac{1}{2}\cdot\dfrac{9}{4}=\dfrac{9}{8}=\boxed{1\dfrac{1}{8}}[/tex]
[tex]b) \ 4\cdot2\dfrac{3}{4}-(1\dfrac{4}{5})^{2}=\not4\cdot\dfrac{11}{\not4}-(\dfrac{9}{5})^{2}=11-\dfrac{81}{25}=11-3\dfrac{6}{25}=10\dfrac{25}{25}-3\dfrac{6}{25}=\boxed{7\dfrac{19}{25}}[/tex]
[tex]c) \ 2\dfrac{1}{4}\cdot(3\dfrac{1}{4}-1\dfrac{7}{12})+(\dfrac{1}{4})^{3}=\dfrac{9}{4}\cdot(3\dfrac{3}{12}-1\dfrac{7}{12})+\dfrac{1}{64}=\dfrac{9}{4}\cdot(2\dfrac{15}{12}-1\dfrac{7}{12})+\dfrac{1}{64}=\\\\=\dfrac{9}{4}\cdot1\dfrac{8}{12}+\dfrac{1}{64}=\dfrac{9}{4}\cdot1\dfrac{2}{3}+\dfrac{1}{64}=\dfrac{\not9^{3}}{4}\cdot\dfrac{5}{\not3}+\dfrac{1}{64}=\dfrac{15}{4}+\dfrac{1}{64}=3\dfrac{3}{4}+\dfrac{1}{64}=\\\\=3\dfrac{48}{64}+\dfrac{1}{64}=\boxed{3\dfrac{49}{64}}[/tex]
