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Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x \ne 1\\
a)E = \dfrac{{x + \sqrt x }}{{x - 2\sqrt x + 1}}:\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} - \dfrac{1}{{1 - \sqrt x }} + \dfrac{{2 - x}}{{x - \sqrt x }}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \sqrt x + 2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{x - 1 + \sqrt x + 2 - x}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}.\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{x}{{\sqrt x - 1}}\\
b)E > 1\\
\Leftrightarrow \dfrac{x}{{\sqrt x - 1}} > 1\\
\Leftrightarrow \dfrac{x}{{\sqrt x - 1}} - 1 > 0\\
\Leftrightarrow \dfrac{{x - \sqrt x + 1}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \sqrt x - 1 > 0\left( {do:x - \sqrt x + 1 > 0} \right)\\
\Leftrightarrow \sqrt x > 1\\
\Leftrightarrow x > 1\\
Vậy\,x > 1\\
c)\\
E = \dfrac{x}{{\sqrt x - 1}}\\
= \dfrac{{x - 1 + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1}}{{\sqrt x - 1}}\\
= \sqrt x + 1 + \dfrac{1}{{\sqrt x - 1}}\\
= \sqrt x - 1 + \dfrac{1}{{\sqrt x - 1}} + 2\\
x > 1\\
\Leftrightarrow \sqrt x - 1 > 0\\
Theo\,Co - si:\\
\sqrt x - 1 + \dfrac{1}{{\sqrt x - 1}} \ge 2\sqrt {\left( {\sqrt x - 1} \right).\dfrac{1}{{\sqrt x - 1}}} \\
\Leftrightarrow \sqrt x - 1 + \dfrac{1}{{\sqrt x - 1}} \ge 2\\
\Leftrightarrow \sqrt x - 1 + \dfrac{1}{{\sqrt x - 1}} + 2 \ge 4\\
\Leftrightarrow E \ge 4\\
\Leftrightarrow GTNN:E = 4\\
Khi:\sqrt x - 1 = \dfrac{1}{{\sqrt x - 1}}\\
\Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 1\\
\Leftrightarrow \sqrt x - 1 = 1\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {tm} \right)\\
d)E = \sqrt x + 1 + \dfrac{1}{{\sqrt x - 1}}\\
E \in Z\\
\Leftrightarrow \left( {\sqrt x - 1} \right) \in \left\{ { - 1;1} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Leftrightarrow x \in \left\{ {0;4} \right\}\\
Do:x > 0\\
\Leftrightarrow x = 4\\
e)E = \dfrac{9}{2}\\
\Leftrightarrow \dfrac{x}{{\sqrt x - 1}} = \dfrac{9}{2}\\
\Leftrightarrow 2x = 9\sqrt x - 9\\
\Leftrightarrow 2x - 9\sqrt x + 9 = 0\\
\Leftrightarrow x - \dfrac{9}{2}\sqrt x + \dfrac{9}{2} = 0\\
\Leftrightarrow x - 2.\sqrt x .\dfrac{9}{4} + \dfrac{{81}}{{16}} - \dfrac{9}{{16}} = 0\\
\Leftrightarrow {\left( {\sqrt x - \dfrac{9}{4}} \right)^2} = \dfrac{9}{{16}}\\
\Leftrightarrow \left( {\sqrt x - \dfrac{9}{4}} \right) = \dfrac{{ \pm 3}}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{3}{4} + \dfrac{9}{4} = 3\\
\sqrt x = - \dfrac{3}{4} + \dfrac{9}{4} = \dfrac{3}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x = \dfrac{9}{4}
\end{array} \right.\left( {tm} \right)
\end{array}$
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