[tex]\it b)\ \ \dfrac{20}{9}\cdot\Big(\dfrac{^{2)}5}{\ 8}+\dfrac{8}{16}\Big)=\dfrac{20}{9}\cdot\dfrac{10+8}{16}=\dfrac{20}{9}\cdot\dfrac{18}{16}=\dfrac{\ 360^{(4}}{144}=\dfrac{\ 90^{(9}}{36}=\dfrac{10^{(2}}{4}=\dfrac{5}{2}\\ \\ Sau:\\ \\ \dfrac{20}{9}\cdot\Big(\dfrac{^{2)}5}{\ 8}+\dfrac{8}{16}\Big)=\dfrac{20}{9}\cdot\dfrac{10+8}{16}=\dfrac{20}{9}\cdot\dfrac{18}{16}=\dfrac{20^{(4}}{16}\cdot\dfrac{18^{(9}}{9}=\dfrac{5}{4}\cdot\dfrac{2}{1}=\dfrac{10^{(2}}{4}=\dfrac{5}{2}[/tex]
Sau:
[tex]\it \dfrac{8^{(2}}{16}=\dfrac{4}{8}[/tex]
Suma din paranteză devine:
[tex]\it \dfrac{5}{8}+\dfrac{4}{8}=\dfrac{9}{8}[/tex]
Acum, exercițiul se poate scrie:
[tex]\it \dfrac{20}{\not9}\cdot\dfrac{\not9}{8}=\dfrac{20^{(4}}{8}=\dfrac{5}{2}=2,5[/tex]
