Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 49
\end{array} \right.\\
P = \left( {\dfrac{3}{{2\sqrt x - 14}} - \dfrac{1}{{2\sqrt x + 14}}} \right):\dfrac{{\sqrt x + 7}}{{x - 49}}\\
= \left( {\dfrac{3}{{2\left( {\sqrt x - 7} \right)}} - \dfrac{1}{{2\left( {\sqrt x + 7} \right)}}} \right).\dfrac{{\left( {\sqrt x - 7} \right)\left( {\sqrt x + 7} \right)}}{{\sqrt x + 7}}\\
= \dfrac{{3.\left( {\sqrt x + 7} \right) - \left( {\sqrt x - 7} \right)}}{{2\left( {\sqrt x - 7} \right)\left( {\sqrt x + 7} \right)}}.\left( {\sqrt x - 7} \right)\\
= \dfrac{{2\sqrt x + 28}}{{2\left( {\sqrt x + 7} \right)}}\\
= \dfrac{{\sqrt x + 14}}{{\sqrt x + 7}}\\
b)P = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 14}}{{\sqrt x + 7}} = \dfrac{3}{2}\\
\Leftrightarrow 3\sqrt x + 21 = 2\sqrt x + 28\\
\Leftrightarrow \sqrt x = 7\\
\Leftrightarrow x = 49\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
c)P = \dfrac{{\sqrt x + 14}}{{\sqrt x + 7}} = \dfrac{{\sqrt x + 7 + 7}}{{\sqrt x + 7}} = 1 + \dfrac{7}{{\sqrt x + 7}}\\
Do:\dfrac{7}{{\sqrt x + 7}} > 0\\
\Leftrightarrow 1 + \dfrac{7}{{\sqrt x + 7}} > 1\\
P - 2 = \dfrac{{\sqrt x + 14}}{{\sqrt x + 7}} - 2\\
= \dfrac{{\sqrt x + 14 - 2\sqrt x - 14}}{{\sqrt x + 7}}\\
= \dfrac{{ - \sqrt x }}{{\sqrt x + 7}} \le 0\left( {do: - \sqrt x \le 0} \right)\\
\Leftrightarrow P \le 2\\
Vậy\,1 < P \le 2
\end{array}$
