Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
N = \left( {\dfrac{{x + 2}}{{x\sqrt x + 1}} - \dfrac{1}{{\sqrt x + 1}}} \right).\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{x + 2 - \left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}}\\
b)N = \dfrac{8}{9}\\
\Leftrightarrow \dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}} = \dfrac{8}{9}\\
\Leftrightarrow 36\sqrt x = 24\left( {x - \sqrt x + 1} \right)\\
\Leftrightarrow 3\sqrt x = 2x - 2\sqrt x + 2\\
\Leftrightarrow 2x - 5\sqrt x + 2 = 0\\
\Leftrightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}\left( {tm} \right)\\
\sqrt x = 2 \Leftrightarrow x = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{1}{4};x = 4\\
c)\dfrac{1}{N} > \dfrac{{3\sqrt x }}{4}\\
\Leftrightarrow \dfrac{{3\left( {x - \sqrt x + 1} \right)}}{{4\sqrt x }} > \dfrac{{3\sqrt x }}{4}\\
\Leftrightarrow x - \sqrt x + 1 > \sqrt x .\sqrt x \\
\Leftrightarrow \sqrt x < 1\\
\Leftrightarrow x < 1\\
Vậy\,0 \le x < 1
\end{array}$
