Đáp án:

$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
P = \frac{{{x^2} - \sqrt x }}{{x + \sqrt x  + 1}} - \frac{{2x + \sqrt x }}{{\sqrt x }} + \frac{{2\left( {x - 1} \right)}}{{\sqrt x  - 1}}\\
 = \frac{{\sqrt x \left( {x\sqrt x  - 1} \right)}}{{x + \sqrt x  + 1}} - \frac{{\sqrt x \left( {2\sqrt x  + 1} \right)}}{{\sqrt x }} + \frac{{2\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\sqrt x  - 1}}\\
 = \frac{{\sqrt x \left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{x + \sqrt x  + 1}} - \left( {2\sqrt x  + 1} \right) + 2\left( {\sqrt x  + 1} \right)\\
 = \sqrt x \left( {\sqrt x  - 1} \right) - 2\sqrt x  - 1 + 2\sqrt x  + 2\\
 = x - \sqrt x  + 1\\
b)x > 0;x \ne 1\\
P = \frac{3}{4}\\
 \Rightarrow x - \sqrt x  + 1 = \frac{3}{4}\\
 \Rightarrow x - \sqrt x  + \frac{1}{4} = 0\\
 \Rightarrow {\left( {\sqrt x } \right)^2} - 2.\sqrt x .\frac{1}{2} + \frac{1}{4} = 0\\
 \Rightarrow {\left( {\sqrt x  - \frac{1}{2}} \right)^2} = 0\\
 \Rightarrow \sqrt x  = \frac{1}{2}\\
 \Rightarrow x = \frac{1}{4}\left( {tmdk} \right)\\
c)P = x - \sqrt x  + 1\\
 = {\left( {\sqrt x  - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4}\forall x > 0;x \ne 1\\
 \Rightarrow Min:P = \frac{3}{4} \Leftrightarrow x = \frac{1}{4}\\
d)\\
Q = \frac{{4\sqrt x }}{{2\left( {x - \sqrt x  + 1} \right)}} = \frac{{2\sqrt x }}{{x - \sqrt x  + 1}}\\
 \Rightarrow Q.x - Q.\sqrt x  + Q = 2\sqrt x \left( {Q > 0} \right)\\
 \Rightarrow Q.x - \left( {Q + 2} \right).\sqrt x  + Q = 0\\
 \Rightarrow \Delta  \ge 0\\
 \Rightarrow {\left( {Q + 2} \right)^2} - 4{Q^2} \ge 0\\
 \Rightarrow  - 3{Q^2} + 4Q + 4 \ge 0\\
 \Rightarrow {Q^2} - \frac{4}{3}Q - \frac{4}{3} \le 0\\
 \Rightarrow \left( {Q - 2} \right)\left( {Q + \frac{2}{3}} \right) \le 0\\
 \Rightarrow 0 < Q \le 2
\end{array}$

Q nguyên chỉ có 2 giá trị Q=1 hoặc Q=2

$\begin{array}{l}
 + Khi:Q = 1\\
 \Rightarrow x - 3\sqrt x  + 1 = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt x  = \frac{{3 + \sqrt 5 }}{2} \Rightarrow x = \frac{{7 + 3\sqrt 5 }}{2}\\
\sqrt x  = \frac{{3 - \sqrt 5 }}{2} \Rightarrow x = \frac{{7 - 3\sqrt 5 }}{2}
\end{array} \right.\\
 + Khi:Q = 2\\
 \Rightarrow 2x - 4\sqrt x  + 2 = 0\\
 \Rightarrow x - 2\sqrt x  + 1 = 0\\
 \Rightarrow {\left( {\sqrt x  - 1} \right)^2} = 0\\
 \Rightarrow \sqrt x  = 1\\
 \Rightarrow x = 1\left( {ktm} \right)
\end{array}$

Vậy có 2 giá trị của x để Q nguyên.