Đáp án:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
b,\\
N = \dfrac{{x - 1}}{{\sqrt x }}\\
c,\\
x > 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x - 1 \ne 0\\
x - \sqrt x \ne 0\\
1 + \sqrt x \ne 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \ne 1\\
\sqrt x \left( {\sqrt x - 1} \right) \ne 0\\
x \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
b,\\
N = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{1}{{x - \sqrt x }}} \right):\left( {\dfrac{1}{{1 + \sqrt x }} + \dfrac{2}{{x - 1}}} \right)\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{{{\sqrt x }^2} - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {\sqrt x - 1} \right) + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}:\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x }}\\
= \dfrac{{x - 1}}{{\sqrt x }}\\
c,\\
N > 0 \Leftrightarrow \dfrac{{x - 1}}{{\sqrt x }} > 0\\
\sqrt x > 0,\,\,\,\forall x > 0,x \ne 1\\
\Rightarrow x - 1 > 0\\
\Rightarrow x > 1
\end{array}\)
