Đáp án:

c. \(MinP = 48\)

Giải thích các bước giải:

\(\begin{array}{l}
a.P = \left[ {\dfrac{{{{\left( {\sqrt x  + 2} \right)}^2} + \sqrt x \left( {2 - \sqrt x } \right) + 4x + 2\sqrt x  - 4}}{{\left( {2 - \sqrt x } \right)\left( {\sqrt x  + 2} \right)}}} \right]:\left[ {\dfrac{{2\sqrt x  - \sqrt x  - 3}}{{\sqrt x \left( {2 - \sqrt x } \right)}}} \right]\\
 = \dfrac{{x + 4\sqrt x  + 4 + 2\sqrt x  - x + 4x + 2\sqrt x  - 4}}{{\left( {2 - \sqrt x } \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\
 = \dfrac{{4x + 8\sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\
 = \dfrac{{4\sqrt x \left( {\sqrt x  + 2} \right)}}{{\left( {2 - \sqrt x } \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x  - 3}}\\
 = \dfrac{{4x}}{{\sqrt x  - 3}}\\
c.P < 0\\
 \to \dfrac{{4x}}{{\sqrt x  - 3}} < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
\sqrt x  - 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
\sqrt x  - 3 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x < 9
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x > 9
\end{array} \right.\left( l \right)
\end{array} \right.\\
 \to 0 < x < 9;x \ne 4\\
c.P = \dfrac{{4x}}{{\sqrt x  - 3}}\\
 \to \dfrac{1}{P} = \dfrac{{\sqrt x  - 3}}{{4x}} = \dfrac{1}{{4\sqrt x }} - \dfrac{3}{{4x}}\\
 =  - \left( {\dfrac{3}{{4x}} - \dfrac{1}{{4\sqrt x }}} \right) =  - \left[ {{{\left( {\dfrac{{\sqrt 3 }}{{2\sqrt x }}} \right)}^2} - 2.\dfrac{{\sqrt 3 }}{{2\sqrt x }}.\dfrac{1}{{4\sqrt 3 }} + {{\left( {\dfrac{1}{{4\sqrt 3 }}} \right)}^2} - \dfrac{1}{{48}}} \right]\\
 =  - {\left( {\dfrac{{\sqrt 3 }}{{2\sqrt x }} - \dfrac{1}{{4\sqrt 3 }}} \right)^2} + \dfrac{1}{{48}}\\
Do:{\left( {\dfrac{{\sqrt 3 }}{{2\sqrt x }} - \dfrac{1}{{4\sqrt 3 }}} \right)^2} \ge 0\\
 \to  - {\left( {\dfrac{{\sqrt 3 }}{{2\sqrt x }} - \dfrac{1}{{4\sqrt 3 }}} \right)^2} \le 0\\
 \to  - {\left( {\dfrac{{\sqrt 3 }}{{2\sqrt x }} - \dfrac{1}{{4\sqrt 3 }}} \right)^2} + \dfrac{1}{{48}} \le \dfrac{1}{{48}}\\
 \to \dfrac{1}{P} \le \dfrac{1}{{48}}\\
 \to P \ge 48\\
 \to MinP = 48\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{{2\sqrt x }} - \dfrac{1}{{4\sqrt 3 }} = 0\\
 \Leftrightarrow \sqrt x  = 6\\
 \to x = 36
\end{array}\)