Đáp án:
$\begin{array}{l}
B1)\\
Dkxd:x \ge 0\\
Q = \dfrac{{2\sqrt x }}{{x - \sqrt x + 1}}\left( {Q \ge 0} \right)\\
\Rightarrow Q.x - Q.\sqrt x + Q = 2\sqrt x \\
\Rightarrow Q.x - \left( {Q + 2} \right).\sqrt x + Q = 0\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {\left( {Q + 2} \right)^2} - 4.Q.Q \ge 0\\
\Rightarrow {Q^2} + 4Q + 4 - 4{Q^2} \ge 0\\
\Rightarrow 3{Q^2} - 4Q - 4 \le 0\\
\Rightarrow \left( {3Q + 2} \right)\left( {Q - 2} \right) \le 0\\
\Rightarrow - \dfrac{2}{3} \le Q \le 2\\
Do:Q \ge 0\\
\Rightarrow 0 \le Q \le 2\\
Vay\,Q \in \left\{ {0;1;2} \right\}\\
B2)\\
A = \dfrac{{\sqrt x + 2}}{{x + \sqrt x + 1}}\left( {Dkxd:x \ge 0} \right)\\
\Rightarrow A.x + A.\sqrt x + A = \sqrt x + 2\\
\Rightarrow A.x + \left( {A - 1} \right).\sqrt x + A - 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta \ge 0\\
\dfrac{{1 - A}}{A} \ge 0\\
\dfrac{{A - 2}}{A} \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\left( {A - 1} \right)^2} - 4.A.\left( {A - 2} \right) \ge 0\\
0 < A \le 1\\
\left[ \begin{array}{l}
A \ge 2\\
A < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{A^2} - 2A + 1 - 4{A^2} + 8A \ge 0\\
A \ge 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3{A^2} - 6A - 1 \le 0\\
A \ge 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{3 - 2\sqrt 3 }}{3} \le A \le \dfrac{{3 + 2\sqrt 3 }}{3}\\
A \ge 2
\end{array} \right.\\
\Rightarrow 2 \le A \le \dfrac{{3 + 2\sqrt 3 }}{3}\\
\Rightarrow A = 2\,khi:x = 0
\end{array}$
