Đáp án:
a. \(\dfrac{{\sqrt x }}{{\sqrt x - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 1\\
P = \left[ {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]:\left[ {\dfrac{{2\left( {\sqrt x + 1} \right) - 2 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right]\\
= \left[ {\dfrac{{x + \sqrt x + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right].\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{2\sqrt x + 2 - 2 + x}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x + 2\sqrt x }}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
b.P > 2\\
\to \dfrac{{\sqrt x }}{{\sqrt x - 1}} > 2\\
\to \dfrac{{\sqrt x - 2\sqrt x + 2}}{{\sqrt x - 1}} > 0\\
\to \dfrac{{2 - \sqrt x }}{{\sqrt x - 1}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2 - \sqrt x > 0\\
\sqrt x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2 - \sqrt x < 0\\
\sqrt x - 1 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
1 < x < 4\\
\left\{ \begin{array}{l}
x > 4\\
x < 1
\end{array} \right.\left( l \right)
\end{array} \right.\\
c.\sqrt P = \sqrt {\dfrac{{2 - \sqrt x }}{{\sqrt x - 1}}} \\
= \sqrt {\dfrac{{ - \left( {\sqrt x - 2} \right)}}{{\sqrt x - 1}}} \\
= \sqrt {\dfrac{{ - \left( {\sqrt x - 1} \right) + 1}}{{\sqrt x - 1}}} \\
= \sqrt { - 1 + \dfrac{1}{{\sqrt x - 1}}}
\end{array}\)
Để \(\sqrt P \) đạt GTNN
⇔ \({\dfrac{1}{{\sqrt x - 1}}}\) đạt GTLN
⇔\({\sqrt x - 1}\) đạt GTNN
\(\begin{array}{l}
\Leftrightarrow \sqrt x - 1 = 1\\
\Leftrightarrow x = 4\\
\to Min\sqrt P = \sqrt { - 1 + 1} = 0
\end{array}\)
