a)
[tex]\it \dfrac{^{3)}3}{\ 4} -\dfrac{5}{12}+\dfrac{^{4)}2}{\ 3}=\dfrac{9-5+8}{12}=\dfrac{12}{12}=1[/tex]
c')
[tex]\it \dfrac{5}{12}:\Big(\dfrac{^{3)}3}{\ 4}-\dfrac{^{4)}1}{\ 3}\Big)-\dfrac{1}{6}:\Big(\dfrac{7}{24}-\dfrac{^{3)}1}{\ 8}\Big)=\dfrac{5}{12}:\dfrac{9-4}{12}-\dfrac{1}{6}:\dfrac{7-3}{24}=\\ \\ \\ =\dfrac{5}{12}:\dfrac{5}{12}-\dfrac{1}{6}:\dfrac{\ 4^{(4}}{24}=1-\dfrac{1}{6}:\dfrac{1}{6}=1-1=0[/tex]
