Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 4;x\# 9\\
a)N = \left( {\dfrac{{x - 3\sqrt x }}{{x - 9}} - 1} \right):\left( {\dfrac{{9 - x}}{{x + \sqrt x - 6}} + \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} - 1} \right)\\
:\dfrac{{9 - x + \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x + 3}} - 1} \right).\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{9 - x + x - 9 - x + 4}}\\
= \dfrac{{\sqrt x - \sqrt x + 3}}{{\sqrt x + 3}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{4 - x}}\\
= \dfrac{3}{1}.\dfrac{{\sqrt x - 2}}{{ - \left( {\sqrt x - 2} \right).\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{ - 3}}{{\sqrt x + 2}}\\
b)N < 1\\
\Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 2}} < 1\\
\Leftrightarrow \dfrac{{ - 3 - \sqrt x - 2}}{{\sqrt x + 2}} < 0\\
\Leftrightarrow \dfrac{{ - 5 - \sqrt x }}{{\sqrt x + 2}} < 0\\
\Leftrightarrow \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}} > 0\left( {tm} \right)
\end{array}$
Vậy N<1 với mọi giá trị của x thỏa mãn $x \ge 0;x\# 4;x\# 9$
$\begin{array}{l}
c)N = \dfrac{{ - 3}}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \left( {\sqrt x + 2} \right) \in U\left( 3 \right)\\
\Leftrightarrow \sqrt x + 2 = 3\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
