Práctica 0 ejercicio 1f - versión 2017

EJERCICIO 1: Calcule

(f) $ \dfrac{ \left( \dfrac{3}{4} \right) \left( \dfrac{8}{9} \right)}{\dfrac{4}{9}}$

Solución:

\begin{eqnarray*}
\dfrac{ \left( \dfrac{3}{4} \right) \left( \dfrac{8}{9} \right)}{\dfrac{4}{9}} & = & \dfrac{ \stackrel{1}{\left( \dfrac{\cancel{3}}{4}\right)}
\left( \dfrac{8}{\cancel{9}_3} \right)}{\dfrac{4}{9}}\\
& = & \dfrac{ \left( \dfrac{1}{\cancel{4}_1} \right)
\stackrel{2}{\left( \dfrac{\cancel{8}}{3}\right)}
}{\dfrac{4}{9}}\\
& = & \cfrac{\frac{ 2}{3}}{\frac{ 4}{9}}= \dfrac{ 2}{3} \cdot \dfrac{ 9}{4} \\
\end{eqnarray*}

\begin{eqnarray*}
& = & \dfrac{ \cancel{2}}{3} \cdot \dfrac{ 9}{\cancel{4}_2} \\
& = & \dfrac{ 1}{\cancel{3}} \cdot \dfrac{\stackrel{3}{\cancel{9}}}{2} \\
& = & \dfrac{ 3}{2}
\end{eqnarray*}


$$\boxed{ \dfrac{ \left( \dfrac{3}{4} \right) \left( \dfrac{8}{9} \right)}{\dfrac{4}{9}}= \dfrac{ 3}{2}}$$