Giải thích các bước giải:
\[\begin{array}{l}
a,\\
D = \left( {\frac{{x\sqrt x }}{{x\sqrt x - 1}} + \frac{{\sqrt x }}{{1 - \sqrt x }}} \right):\left( {\frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}} \right)\\
\Leftrightarrow D = \left( {\frac{{x\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{{\sqrt x }}{{\sqrt x - 1}}} \right).\frac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
\Leftrightarrow D = \left( {\frac{{x\sqrt x - \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right).\frac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
\Leftrightarrow D = \left( {\frac{{ - \sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}} \right).\frac{1}{{\sqrt x + 1}}\\
\Leftrightarrow D = \frac{{\sqrt x }}{{1 - \sqrt x }}
\end{array}\]
b,
Thay x=25 vào D ta được D=-5/4
c,
\[\begin{array}{l}
D > 1 \Leftrightarrow \frac{{\sqrt x }}{{1 - \sqrt x }} > 1\\
\Leftrightarrow \frac{{\sqrt x }}{{1 - \sqrt x }} - 1 > 0\\
\Leftrightarrow \frac{{2\sqrt x - 1}}{{1 - \sqrt x }} > 0\\
\Leftrightarrow \frac{{2\sqrt x - 1}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \frac{1}{2} < \sqrt x < 1\\
\Leftrightarrow \frac{1}{4} < x < 1
\end{array}\]
d,\[D = \frac{{\sqrt x }}{{1 - \sqrt x }} = \frac{{\left( {\sqrt x - 1} \right) + 1}}{{1 - \sqrt x }} = - 1 + \frac{1}{{1 - \sqrt x }}\]
để D là số nguyên thì \[\left( {1 - \sqrt x } \right) \in U(1) = \left\{ { \pm 1} \right\} \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\]
