Vom simplifica:
[tex]a) \ \dfrac{\not7}{5} \cdot \dfrac{3}{\not49} = \dfrac{1}{5} \cdot \dfrac{3}{7} = \bf \dfrac{3}{35}[/tex]
[tex]b) \ \not2 \cdot \dfrac{7}{\not8} = 1 \cdot \dfrac{7}{4} = \bf \dfrac{7}{4}[/tex]
[tex]c) \ 4 : \dfrac{6}{5} = \not4 \cdot \dfrac{5}{\not6} = 2 \cdot \dfrac{5}{3} = \bf \dfrac{10}{3}[/tex]
[tex]d) \ \dfrac{2}{9} : \dfrac{7}{3} = \dfrac{2}{\not9} \cdot \dfrac{\not3}{7} = \dfrac{2}{3} \cdot \dfrac{1}{7} = \bf \dfrac{2}{21}[/tex]
[tex]e) \ \dfrac{16}{3} : \bigg(\dfrac{5}{2} : \dfrac{15}{4} \bigg) = \dfrac{16}{3} : \bigg(\dfrac{\not5}{\not2} \cdot \dfrac{\not4}{\not15} \bigg) = \dfrac{16}{3} : \bigg(\dfrac{1}{1} \cdot \dfrac{2}{3} \bigg) = \dfrac{16}{3} : \dfrac{2}{3} = \dfrac{\not16}{\not3} \cdot \dfrac{\not3}{\not2} = \bf8[/tex]
[tex]f) \ \bigg(\dfrac{2}{3} \bigg)^3 = \dfrac{2^3}{3^3} = \bf \dfrac{8}{27}[/tex]
[tex]g) \ \bigg(\dfrac{1}{4} \bigg)^2 \cdot \bigg(\dfrac{2}{5} \bigg)^2 = \dfrac{1}{4^2} \cdot \dfrac{2^2}{5^2} = \dfrac{1}{4^2} \cdot \dfrac{4}{25} = \dfrac{1}{4} \cdot \dfrac{1}{25} = \bf \dfrac{1}{100}[/tex]
[tex]h) \ \bigg(\dfrac{1}{3} \bigg)^2 - \dfrac{2}{3} = \dfrac{1}{3^2} - \dfrac{^{3)} 2}{3} = \dfrac{1}{9} - \dfrac{6}{9} = \bf - \dfrac{5}{9}[/tex]
[tex]i) \ \bigg(\dfrac{3}{7} \bigg)^2 : \dfrac{5}{7} = \dfrac{3^2}{7^2} \cdot \dfrac{7}{5} = \dfrac{9}{7} \cdot \dfrac{1}{5} = \bf \dfrac{9}{35}[/tex]
(verifică la h, dacă este scădere sau înmulțire)
