Đáp án:

\(\begin{array}{l}
4,\\
x \in \left\{ {0;\dfrac{1}{4};4} \right\}\\
5,\\
{P_{\min }} = 10 \Leftrightarrow x = 25
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
4,\\
P = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}}\,\,\,\,\,\,\left( {x \ge 0} \right)\\
P + 2 = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} + 2 = \dfrac{{\left( {2\sqrt x  - 1} \right) + 2\left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}\\
 = \dfrac{{2\sqrt x  - 1 + 2\sqrt x  + 2}}{{\sqrt x  + 1}} = \dfrac{{4\sqrt x  + 1}}{{\sqrt x  + 1}}\\
x \ge 0 \Rightarrow \left\{ \begin{array}{l}
4\sqrt x  + 1 \ge 1 > 0\\
\sqrt x  + 1 \ge 1 > 0
\end{array} \right. \Rightarrow \dfrac{{4\sqrt x  + 1}}{{\sqrt x  + 1}} > 0\\
 \Rightarrow P + 2 > 0,\,\,\,\forall x \ge 0\\
 \Rightarrow P >  - 2,\,\,\,\forall x \ge 0\\
P - 2 = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} - 2 = \dfrac{{\left( {2\sqrt x  - 1} \right) - 2.\left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}\\
 = \dfrac{{2\sqrt x  - 1 - 2\sqrt x  - 2}}{{\sqrt x  + 1}} = \dfrac{{ - 3}}{{\sqrt x  + 1}}\\
x \ge 0 \Rightarrow \sqrt x  + 1 \ge 1 > 0 \Rightarrow \dfrac{{ - 3}}{{\sqrt x  + 1}} < 0\\
 \Rightarrow P - 2 < 0,\,\,\,\forall x \ge 0\\
 \Leftrightarrow P < 2,\,\,\,\forall x \ge 0\\
 \Rightarrow  - 2 < P < 2,\,\,\,\forall x \ge 0\\
P \in Z \Rightarrow P \in \left\{ { - 1;0;1} \right\}\\
*)\\
P =  - 1 \Rightarrow \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} =  - 1\\
 \Leftrightarrow 2\sqrt x  - 1 =  - \sqrt x  - 1\\
 \Leftrightarrow 2\sqrt x  + \sqrt x  = 0\\
 \Leftrightarrow 3\sqrt x  = 0\\
 \Leftrightarrow \sqrt x  = 0\\
 \Leftrightarrow x = 0\\
*)\\
P = 0 \Leftrightarrow \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} = 0\\
 \Leftrightarrow 2\sqrt x  - 1 = 0\\
 \Leftrightarrow \sqrt x  = \dfrac{1}{2}\\
 \Leftrightarrow x = \dfrac{1}{4}\\
*)\\
P = 1 \Leftrightarrow \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} = 1\\
 \Leftrightarrow 2\sqrt x  - 1 = \sqrt x  + 1\\
 \Leftrightarrow \sqrt x  - 1 = 1\\
 \Leftrightarrow \sqrt x  = 2\\
 \Leftrightarrow x = 4\\
Vậy\,\,\,x \in \left\{ {0;\dfrac{1}{4};4} \right\}\\
5,\\
Q = \dfrac{{x + 15}}{{\sqrt x  - 1}}\\
Q - 10 = \dfrac{{x + 15}}{{\sqrt x  - 1}} - 10\\
 = \dfrac{{x + 15 - 10\left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}}\\
 = \dfrac{{x + 15 - 10\sqrt x  + 10}}{{\sqrt x  - 1}}\\
 = \dfrac{{x - 10\sqrt x  + 25}}{{\sqrt x  - 1}}\\
 = \dfrac{{x - 2.\sqrt x .5 + {5^2}}}{{\sqrt x  - 1}}\\
 = \dfrac{{{{\left( {\sqrt x  - 5} \right)}^2}}}{{\sqrt x  - 1}}\\
{\left( {\sqrt x  - 5} \right)^2} \ge 0,\,\,\,\forall x > 1\\
\sqrt x  - 1 > 0,\,\,\,\forall x > 1\\
 \Rightarrow \dfrac{{{{\left( {\sqrt x  - 5} \right)}^2}}}{{\sqrt x  - 1}} \ge 0,\,\,\,\forall x > 1\\
 \Rightarrow P - 10 \ge 0,\,\,\,\forall x > 1\\
 \Leftrightarrow P \ge 10,\,\,\,\forall x > 1\\
 \Rightarrow {P_{\min }} = 10 \Leftrightarrow {\left( {\sqrt x  - 5} \right)^2} = 0 \Leftrightarrow \sqrt x  = 5 \Leftrightarrow x = 25\\
Vậy \,\,\,{P_{\min }} = 10 \Leftrightarrow x = 25
\end{array}\)