Đáp án:

 c) \(0 \le x < 1\)

Giải thích các bước giải:

\(\begin{array}{l}
a)Thay:x = 16\\
 \to \sqrt x  = 4\\
 \to B = \dfrac{{4 + 3}}{{16 + 4 + 1}} = \dfrac{7}{{21}} = \dfrac{1}{3}\\
b)DK:x \ge 0;x \ne 1\\
A = \dfrac{{\left( {2x + 1} \right)\left( {\sqrt x  + 1} \right) - x\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{2x\sqrt x  + 2x + \sqrt x  + 1 - x\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{x\sqrt x  + 2x + \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
P = A:B = \dfrac{{x\sqrt x  + 2x + \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  + 3}}{{x + \sqrt x  + 1}}\\
 = \dfrac{{x\sqrt x  + 2x + \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{x + \sqrt x  + 1}}{{\sqrt x  + 3}}\\
 = \dfrac{{x\sqrt x  + 2x + \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
c)P < \dfrac{1}{2}\\
 \to \dfrac{{x\sqrt x  + 2x + \sqrt x  + 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} < \dfrac{1}{2}\\
 \to \dfrac{{2x\sqrt x  + 4x + 2\sqrt x  + 4 - \left( {x - 1} \right)\left( {\sqrt x  + 3} \right)}}{{2\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} < 0\\
 \to \dfrac{{2x\sqrt x  + 4x + 2\sqrt x  + 4 - x\sqrt x  - 3x + \sqrt x  + 3}}{{2\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} < 0\\
 \to \dfrac{{x\sqrt x  + x + 3\sqrt x  + 7}}{{2\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} < 0\\
 \to \dfrac{{x\sqrt x  + x + 3\sqrt x  + 7}}{{\sqrt x  - 1}} < 0\left( {do:\left\{ \begin{array}{l}
\sqrt x  + 1 > 0\\
\sqrt x  + 3 > 0
\end{array} \right.\forall x \ge 0} \right)\\
Mà:x\sqrt x  + x + 3\sqrt x  + 7 > 0\forall x \ge 0\\
 \to \sqrt x  - 1 < 0\\
 \to x < 1\\
 \to 0 \le x < 1
\end{array}\)