[tex]\bf a)\; 4\dfrac{1}{2} : \dfrac{2}{3} = \dfrac{4 \cdot 2 +1}{2} :\dfrac{2}{3} = \dfrac{9}{2} :\dfrac{2}{3} =\dfrac{9}{2} \cdot \dfrac{3}{2} = \boxed{\bf\dfrac{27}{4} }[/tex]
[tex]\bf \;[/tex]
[tex]\bf b)\; \dfrac{7}{12} : 3\dfrac{1}{16} =\dfrac{7}{12} : \dfrac{3 \cdot 16 +1 }{16} =\dfrac{7}{12} : \dfrac{49}{16} = \bigg( \dfrac{7}{12}\cdot \dfrac{16}{49}\bigg) ^{(28} = \boxed{\bf \dfrac{4}{21} }[/tex]
[tex]\bf \;[/tex]
[tex]\bf c) \; 5 : 1\dfrac{1}{4} = 5: \dfrac{1\cdot 4 +1}{4} = 5:\dfrac{5}{4} = 5 \cdot \dfrac{4}{5} = \boxed{\bf 4}[/tex]
[tex]\bf \;[/tex]
[tex]\bf d)\; 2\dfrac{2}{3} : 2\dfrac{1}{3 } = \dfrac{2 \cdot 3 +2}{3} : \dfrac{2\cdot 3 +1}{3} = \dfrac{8}{3} : \dfrac{7}{3} = \bigg( \dfrac{8}{3} \cdot \dfrac{3}{7 } \bigg) ^{(3} = \boxed{ \bf \dfrac{8}{7} }[/tex]
[tex]\bf \;[/tex]
[tex]\bf e)\; 4 \dfrac{1}{2} : 3\dfrac{1}{2} = \dfrac{4\cdot 2 +1}{2} : \dfrac{3\cdot 2 +1}{2} =\dfrac{9}{2} : \dfrac{7}{2} = \bigg( \dfrac{9}{2} \cdot \dfrac{2}{7} \bigg) ^{(2} = \boxed{\bf \dfrac{9}{7} }[/tex]
[tex]\bf \;[/tex]
[tex]\bf f)\; 16 : 9 \dfrac{3}{5} = 16: \dfrac{9 \cdot 5 +3}{5} = 16 : \dfrac{48}{5} = \bigg(16 \cdot \dfrac{5}{48} \bigg) ^{(16} = \boxed{ \bf \dfrac{5}{3} }[/tex]
[tex]\bf \;[/tex]
[tex]\bf g)\; 3\dfrac{3}{4} : 1 \dfrac{1}{4} = \dfrac{3 \cdot 4 +3 }{4 } : \dfrac{1\cdot 4 +1}{4} = \dfrac{15}{4} : \dfrac{5}{4} = \bigg( \dfrac{15}{4} \cdot \dfrac{4}{5} \bigg) ^{(4}= \dfrac{15}{5} =\boxed{\bf 3}[/tex]
[tex]\bf \;[/tex]
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[tex]\bf a)~~ 4\dfrac{1}{2} : \dfrac{2}{3} = \dfrac{4 \cdot 2 +1}{2} :\dfrac{2}{3} = \dfrac{9}{2} :\dfrac{2}{3} =\dfrac{9}{2} \cdot \dfrac{3}{2} = \boxed{\boxed{\bf\dfrac{27}{4} }}[/tex]
[tex]\it~~[/tex]
[tex]\bf b)~\dfrac{7}{12}:3\dfrac{1}{16} =\dfrac{7}{12}:\dfrac{3 \cdot 16 +1 }{16} =\dfrac{7}{12} : \dfrac{49}{16} =\dfrac{\not7}{\not12}\cdot \dfrac{\not16}{\not49}=\dfrac{1}{3}\cdot \dfrac{4}{7}=\boxed{\boxed{\bf \dfrac{4}{21}}}[/tex]
[tex]\it~~~[/tex]
[tex]\bf c)~ 5 : 1\dfrac{1}{4} = 5: \dfrac{1\cdot 4 +1}{4} = 5:\dfrac{5}{4} = 5 \cdot \dfrac{4}{5} = \not5 \cdot \dfrac{4}{\not5} =\boxed{\boxed{\bf 4}}[/tex]
[tex]\it~~~[/tex]
[tex]\bf d)~ 2\dfrac{2}{3} : 2\dfrac{1}{3 } = \dfrac{2 \cdot 3 +2}{3} : \dfrac{2\cdot 3 +1}{3} = \dfrac{8}{3} : \dfrac{7}{3} = \dfrac{8}{\not3} \cdot \dfrac{\not3}{7} =\boxed{\boxed{ \bf \dfrac{8}{7}}}[/tex]
[tex]\it~~[/tex]
[tex]\bf e)~~4 \dfrac{1}{2} : 3\dfrac{1}{2} = \dfrac{4\cdot 2 +1}{2} : \dfrac{3\cdot 2 +1}{2} =\dfrac{9}{2} : \dfrac{7}{2} =\dfrac{9}{\not2} \cdot \dfrac{\not2}{7} = \boxed{\boxed{\bf \dfrac{9}{7}}}[/tex]
[tex]\it~~[/tex]
[tex]\bf f)~~ 16 : 9 \dfrac{3}{5} = 16: \dfrac{9 \cdot 5 +3}{5} = 16 : \dfrac{48}{5} =\not 16 \cdot \dfrac{5}{\not48} =1 \cdot \dfrac{5}{3} =\boxed{ \boxed{ \bf \dfrac{5}{3}}}[/tex]
[tex]\it~~[/tex]
[tex]\bf g)~~ 3\dfrac{3}{4} : 1 \dfrac{1}{4} = \dfrac{3 \cdot 4 +3 }{4 } : \dfrac{1\cdot 4 +1}{4} = \dfrac{15}{4} : \dfrac{5}{4} = \dfrac{\not15}{\not4} \cdot \dfrac{\not4}{\not5}= \dfrac{3}{1} \cdot \dfrac{1}{1}= \boxed{\boxed{\bf 3} }[/tex]
[tex]\it~~[/tex]