Efectuăm separat parantezele rotunde:
[tex]\it \Big(\dfrac{1}{2}\Big)^3=\dfrac{1}{8}\\ \\ \\ \Big(\dfrac{^{4)}2}{\ 3}-\dfrac{5}{12}\Big)^2= \Big(\dfrac{\ 3^{(3}}{12}\Big)^2= \Big(\dfrac{1}{4}\Big)^2=\dfrac{1}{16}\\ \\ \\ \dfrac{93}{90}-\dfrac{^{2)}39}{\ 45}= \dfrac{93-78}{90}=\dfrac{\ \ 15^{(15}}{90}=\dfrac{1}{6}[/tex]
Acum, exercițiul devine:
[tex]\it 1-\Big[\dfrac{1}{8}+\Big(\dfrac{1}{16}+\dfrac{1}{6}-\dfrac{1}{6}\Big):\dfrac{3}{2}\Big]\cdot6= 1-\Big(\dfrac{1}{8}+\dfrac{1}{16}\cdot\dfrac{2}{3}\Big)\cdot6=\\ \\ \\ =1-\Big(\dfrac{^{3)}1}{\ 8}+\dfrac{1}{24}\Big)\cdot6 = 1- \dfrac{4}{24}\cdot6=1-\dfrac{24}{24}=1-1=0[/tex]
