Đáp án:
$\begin{array}{l}
12)a)Dkxd:x > 0;x\# 4\\
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 2}} + \dfrac{{4\sqrt x - 3}}{{2\sqrt x - x}}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x }} - \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}}} \right)\\
= \dfrac{{\sqrt x .\sqrt x - 4\sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \sqrt x \left( {\sqrt x - 4} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4\sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x - 4 - x + 4\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}{{4\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 3}}{4}\\
b)P > 0\\
\Leftrightarrow \dfrac{{\sqrt x - 3}}{4} > 0\\
\Leftrightarrow \sqrt x - 3 > 0\\
\Leftrightarrow \sqrt x > 3\\
\Leftrightarrow x > 9\\
Vay\,x > 9\\
c)\sqrt P = \sqrt {\dfrac{{\sqrt x - 3}}{4}} = \dfrac{{\sqrt {\sqrt x - 3} }}{2} \ge 0\\
\Leftrightarrow GTNN:\sqrt P = 0\\
Khi:x = 9\\
13)\\
a)Dkxd:x \ge 0;x\# 1\\
A = \left( {\dfrac{1}{{1 - \sqrt x }} + \dfrac{1}{{1 + \sqrt x }}} \right):\left( {\dfrac{1}{{1 - \sqrt x }} - \dfrac{1}{{1 + \sqrt x }}} \right) + \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{1 + \sqrt x + 1 - \sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}:\dfrac{{1 + \sqrt x - 1 + \sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{2}{{1 - x}}.\dfrac{{1 - x}}{{2\sqrt x }} + \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{1}{{\sqrt x }} + \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{1 - \sqrt x + \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \dfrac{1}{{\sqrt x - x}}\\
b)x = 7 + 4\sqrt 3 \left( {tmdk} \right) = {\left( {2 + \sqrt 3 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 + \sqrt 3 \\
\Leftrightarrow A = \dfrac{1}{{2 + \sqrt 3 - 7 - 4\sqrt 3 }}\\
= \dfrac{1}{{ - 5 - 2\sqrt 3 }}\\
= \dfrac{{2\sqrt 3 - 5}}{{13}}\\
c)A = \dfrac{1}{{\sqrt x - x}}\\
\sqrt x - x = - \left( {x - \sqrt x + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
= - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{{\sqrt x - x}} \ge 4\\
\Leftrightarrow A \ge 4\\
\Leftrightarrow GTNN:A = 4\,khi:x = \dfrac{1}{4}
\end{array}$
