Đáp án:
$\begin{array}{l}
b)B = \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}\left( {x > 0} \right)\\
\Rightarrow B.x + B.\sqrt x + B = \sqrt x \\
\Rightarrow B.x + \left( {B - 1} \right).\sqrt x + B = 0\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{1 - B}}{B} \ge 0\\
\Delta \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
{B^2} - 2B + 1 - 4{B^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
3{B^2} + 2B - 1 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
\left( {3B - 1} \right)\left( {B + 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
- 1 \le B \le \dfrac{1}{3}
\end{array} \right.\\
\Rightarrow 0 \le B \le \dfrac{1}{3}\\
\Rightarrow GTLN:B = \dfrac{1}{3}\,khi:x = 1\\
c)C = \dfrac{{\sqrt x }}{{x - \sqrt x + 1}}\\
\Rightarrow C.x - C.\sqrt x + C = \sqrt x \\
\Rightarrow C.x - \left( {C + 1} \right).\sqrt x + C = 0\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
\Delta \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
{\left( {C + 1} \right)^2} - 4{C^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
{C^2} + 2C + 1 - 4{C^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
3{C^2} - 2C - 1 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
\left( {3C + 1} \right)\left( {C - 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
- \dfrac{1}{3} \le C \le 1
\end{array} \right.\\
\Rightarrow 0 < C \le 1\\
\Rightarrow GTLN:C = 1\,khi:x = 1
\end{array}$
