Đáp án:
$\begin{array}{l}
1)Dk:x > 0;x \ne 4\\
A = \left( {\dfrac{{\sqrt x - 1}}{{x - 4}} - \dfrac{{\sqrt x + 1}}{{x + 4\sqrt x + 4}}} \right):\dfrac{{x\sqrt x }}{{{{\left( {4 - x} \right)}^2}}}\\
= \left( {\dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}} \right).\dfrac{{{{\left( {x - 4} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 2} \right) - \left( {\sqrt x + 1} \right).\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right).{{\left( {\sqrt x + 2} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x - 2} \right)}^2}.{{\left( {\sqrt x + 2} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{x + \sqrt x - 2 - \left( {x - \sqrt x - 2} \right)}}{1}.\dfrac{{\sqrt x - 2}}{{x\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{x\sqrt x }}.\left( {\sqrt x - 2} \right)\\
= \dfrac{{2\left( {\sqrt x - 2} \right)}}{x}\\
2)x > 0;x \ne 4\\
x = 4 + 2\sqrt 3 \left( {tmdk} \right)\\
\Rightarrow x = {\left( {\sqrt 3 + 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 3 + 1\\
\Rightarrow A = \dfrac{{2.\left( {\sqrt 3 + 1 - 2} \right)}}{{4 + 2\sqrt 3 }}\\
= \dfrac{{\sqrt 3 - 1}}{{2 + \sqrt 3 }}\\
= \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}}\\
= - 5 + 3\sqrt 3 \\
3)A \ge \dfrac{1}{4}\\
\Rightarrow \dfrac{{2\left( {\sqrt x - 2} \right)}}{x} \ge \dfrac{1}{4}\\
\Rightarrow \dfrac{{8\sqrt x - 16 - x}}{{4x}} \ge 0\\
\Rightarrow - x + 8\sqrt x - 16 \ge 0\left( {do:x > 0} \right)\\
\Rightarrow x - 8\sqrt x + 16 \le 0\\
\Rightarrow {\left( {\sqrt x - 4} \right)^2} \le 0\\
\Rightarrow \sqrt x - 4 = 0\left( {do:{{\left( {\sqrt x - 4} \right)}^2} \ge 0\forall x} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow x = 16\left( {tm} \right)
\end{array}$
Vậy x=16
