Respuestas

2012-04-26T04:39:41+02:00

\begin{gathered} \sqrt x - \sqrt[4]{x} = 2 \hfill \\ \sqrt x - 2 = \sqrt[4]{x}/\left( {} \right)^2 \hfill \\ \left( {\sqrt x - 2} \right)^2 = \left( {\sqrt[4]{x}} \right)^2 \hfill \\ x + 4 - 4\sqrt x = \sqrt x / + 4\sqrt x \hfill \\ x + 4 = 5\sqrt x /\left( {} \right)^2 \hfill \\ \left( {x + 4} \right)^2 = 25x \hfill \\ x^2 + 8x + 16 = 25x \hfill \\ x^2 - 17x + 16 = 0 \hfill \\ \left( {x - 16} \right)\left( {x - 1} \right) = 0 \hfill \\ \end{gathered}

 

\Rightarrow x = 16 \vee x = 1

 

Pero x=1, no es solución

 

así que S={16}