Giải thích các bước giải:
\(\begin{array}{l}
5,\\
a,\\
DK:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x - 1 \ne 0\\
\sqrt x - x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
P = \dfrac{x}{{\sqrt x - 1}} + \dfrac{{2x - \sqrt x }}{{\sqrt x - x}}\\
= \dfrac{x}{{\sqrt x - 1}} + \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)}}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \dfrac{x}{{\sqrt x - 1}} + \dfrac{{2\sqrt x - 1}}{{1 - \sqrt x }}\\
= \dfrac{{x - \left( {2\sqrt x - 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}}\\
= \sqrt x - 1\\
b,\\
x = \dfrac{1}{4} \Rightarrow \sqrt x = \dfrac{1}{2} \Rightarrow P = \sqrt x - 1 = \dfrac{1}{2} - 1 = - \dfrac{1}{2}\\
6,\\
a,\\
DK:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - 1 \ne 0\\
\sqrt x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
A = \dfrac{{x\sqrt x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{x - \sqrt x + 1}}{{\sqrt x - 1}} - \left( {\sqrt x - 1} \right)\\
= \dfrac{{\left( {x - \sqrt x + 1} \right) - {{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {x - \sqrt x + 1} \right) - \left( {x - 2\sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
b,\\
x = \dfrac{1}{4} \Rightarrow \sqrt x = \dfrac{1}{2}\\
\Rightarrow A = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2} - 1}} = - 1\\
c,\\
A < 0 \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 1}} < 0 \Leftrightarrow \sqrt x - 1 < 0 \Leftrightarrow 0 \le x < 1\\
d,\\
\left| A \right| = A \Leftrightarrow A > 0 \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 1}} > 0 \Leftrightarrow \sqrt x - 1 > 0 \Leftrightarrow x > 1
\end{array}\)
