Đáp án:
a) $Q = \dfrac{{\sqrt x + 1}}{{x - 4}}$ với $x \ge 0;x \ne \left\{ {4;9} \right\}$
b) $x \in \left[ {0;\dfrac{1}{4}} \right]$
c) $MinA = - 6 \Leftrightarrow x = 0$
Giải thích các bước giải:
a) ĐKXĐ: $x \ge 0;x \ne \left\{ {4;9} \right\}$
Ta có:
$\begin{array}{l}
Q = \left( {\dfrac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{2 - \sqrt x }} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}} \right):\left( {2 - \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right)\\
= \left( {\dfrac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} + \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}} \right):\left( {\dfrac{{2\left( {\sqrt x + 1} \right) - \sqrt x }}{{\sqrt x + 1}}} \right)\\
= \left( {\dfrac{{\sqrt x + 2 + \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}} \right):\dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{x - 4}}
\end{array}$
Vậy $Q = \dfrac{{\sqrt x + 1}}{{x - 4}}$ với $x \ge 0;x \ne \left\{ {4;9} \right\}$
b) Ta có:
$\begin{array}{l}
\dfrac{1}{Q} \le \dfrac{{ - 5}}{2}\\
\Leftrightarrow \dfrac{{x - 4}}{{\sqrt x + 1}} \le \dfrac{{ - 5}}{2}\\
\Leftrightarrow 2\left( {x - 4} \right) + 5\left( {\sqrt x + 1} \right) \le 0\left( {do:\sqrt x + 1 > 0,\forall x \ge 0} \right)\\
\Leftrightarrow 2x + 5\sqrt x - 3 \le 0\\
\Leftrightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 3} \right) \le 0\\
\Leftrightarrow 2\sqrt x - 1 \le 0\left( {do:\sqrt x + 3 > 0,\forall x \ge 0} \right)\\
\Leftrightarrow \sqrt x \le \dfrac{1}{2}\\
\Leftrightarrow x \le \dfrac{1}{4}
\end{array}$
Vậy $x \in \left[ {0;\dfrac{1}{4}} \right]$ thỏa mãn đề
c) Ta có:
$\begin{array}{l}
A = \dfrac{P}{Q} = \dfrac{{\dfrac{3}{{\sqrt x + 2}}}}{{\dfrac{{\sqrt x + 1}}{{x - 4}}}} = \dfrac{{3\left( {x - 4} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{{3\sqrt x - 6}}{{\sqrt x + 1}}\\
\Rightarrow A + 6 = \dfrac{{3\sqrt x - 6}}{{\sqrt x + 1}} + 6 = \dfrac{{3\sqrt x - 6 + 6\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}} = \dfrac{{9\sqrt x }}{{\sqrt x + 1}} \ge 0,\forall x\\
\Rightarrow A \ge - 6
\end{array}$
Dấu bằng xảy ra $ \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0$
Vậy $MinA = - 6 \Leftrightarrow x = 0$
