Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\\
a,\\
x = 16 \Rightarrow A = \dfrac{{ - 3\sqrt {16} + 1}}{{\sqrt {16} - 3}} = \dfrac{{ - 3.4 + 1}}{{4 - 3}} = - 11\\
b,\\
B = \dfrac{{3\sqrt x - 2}}{{x - 5\sqrt x + 6}} - \dfrac{1}{{\sqrt x - 2}} + \dfrac{{3\sqrt x - 2}}{{3 - \sqrt x }}\\
= \dfrac{{3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{1}{{\sqrt x - 2}} - \dfrac{{3\sqrt x - 2}}{{\sqrt x - 3}}\\
= \dfrac{{\left( {3\sqrt x - 2} \right) - 1.\left( {\sqrt x - 3} \right) - \left( {3\sqrt x - 2} \right).\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 2 - \sqrt x + 3 - \left( {3x - 8\sqrt x + 4} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 2 - \sqrt x + 3 - 3x + 8\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3x + 10\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= - \dfrac{{3x - 10\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= - \dfrac{{\left( {3\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= - \dfrac{{3\sqrt x - 1}}{{\sqrt x - 2}}\\
c,\\
B > - 3\\
\Leftrightarrow - \dfrac{{3\sqrt x - 1}}{{\sqrt x - 2}} > - 3\\
\Leftrightarrow \dfrac{{3\sqrt x - 1}}{{\sqrt x - 2}} < 3\\
\Leftrightarrow \dfrac{{3\sqrt x - 1}}{{\sqrt x - 2}} - 3 < 0\\
\Leftrightarrow \dfrac{{\left( {3\sqrt x - 1} \right) - 3.\left( {\sqrt x - 2} \right)}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \dfrac{{3\sqrt x - 1 - 3\sqrt x + 6}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \dfrac{5}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \sqrt x - 2 < 0\\
\Leftrightarrow 0 < x < 4\\
d,\\
P = \dfrac{A}{B} = A:B = \dfrac{{ - 3\sqrt x + 1}}{{\sqrt x - 3}}:\left( { - \dfrac{{3\sqrt x - 1}}{{\sqrt x - 2}}} \right)\\
= \dfrac{{ - 3\sqrt x + 1}}{{\sqrt x - 3}}:\dfrac{{ - 3\sqrt x + 1}}{{\sqrt x - 2}} = \dfrac{{ - 3\sqrt x + 1}}{{\sqrt x - 3}}.\dfrac{{\sqrt x - 2}}{{ - 3\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}}\\
P - 1 = \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} - 1 = \dfrac{{\left( {\sqrt x - 2} \right) - \left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}} = \dfrac{1}{{\sqrt x - 3}}\\
x > 9 \Leftrightarrow \sqrt x > 3 \Leftrightarrow \sqrt x - 3 > 0\\
\Rightarrow P - 1 = \dfrac{1}{{\sqrt x - 3}} > 0\\
\Rightarrow P > 1,\,\,\,\forall x > 9
\end{array}\)
