Đáp án:

\(\begin{array}{l}
a)x > 0;x \ne 1\\
b)F = \dfrac{{2x + 2}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} = \dfrac{{2x + 2}}{{x + \sqrt x }}\\
c)x \in \emptyset 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x > 0;x \ne 1\\
b)F = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2} + {{\left( {\sqrt x  - 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
 = \dfrac{{x + 2\sqrt x  + 1 + x - 2\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
 = \dfrac{{2x + 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
 = \dfrac{{2x + 2}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} = \dfrac{{2x + 2}}{{x + \sqrt x }}\\
c)F = \dfrac{{2x + 2}}{{x + \sqrt x }} = \dfrac{{2\left( {x + \sqrt x } \right) - 2\sqrt x  + 2}}{{x + \sqrt x }}\\
 = \dfrac{{2\left( {x + \sqrt x } \right) - 2\left( {\sqrt x  + 1} \right) + 4}}{{x + \sqrt x }}\\
 = 2 - \dfrac{2}{{\sqrt x }} + \dfrac{4}{{x + \sqrt x }}\\
F \in Z \to \left\{ \begin{array}{l}
\dfrac{2}{{\sqrt x }} \in Z\\
\dfrac{4}{{x + \sqrt x }} \in Z
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\sqrt x  \in U\left( 2 \right)\\
x + \sqrt x  \in U\left( 4 \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\sqrt x  = 2\\
\sqrt x  = 1
\end{array} \right.\\
\left[ \begin{array}{l}
x + \sqrt x  = 4\\
x + \sqrt x  = 2\\
x + \sqrt x  = 1
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 4\\
x = 1\left( l \right)
\end{array} \right.\\
\left[ \begin{array}{l}
\sqrt x  = \dfrac{{ - 1 \pm \sqrt {17} }}{3}\left( l \right)\\
\sqrt x  = 1\left( l \right)\\
\sqrt x  =  - 2\left( l \right)\\
\sqrt x  = \dfrac{{ - 1 \pm \sqrt 5 }}{3}\left( l \right)
\end{array} \right.
\end{array} \right.\\
 \to x \in \emptyset 
\end{array}\)