Đáp án:

$\begin{array}{l}
1)a)x \ge 0;x \ne 4\\
P = \left( {3 - \dfrac{{3\sqrt x  + 3}}{{\sqrt x  + 2}}} \right):\left( {\dfrac{2}{{\sqrt x  + 2}} + \dfrac{3}{{\sqrt x  - 2}} + \dfrac{{2\sqrt x  - 1}}{{4 - x}}} \right)\\
 = \dfrac{{3\sqrt x  + 6 - 3\sqrt x  - 3}}{{\sqrt x  + 2}}:\\
\dfrac{{2\left( {\sqrt x  - 2} \right) + 3\left( {\sqrt x  + 2} \right) - 2\sqrt x  + 1}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{3}{{\sqrt x  + 2}}.\dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}{{2\sqrt x  - 4 + 3\sqrt x  + 6 - 2\sqrt x  + 1}}\\
 = \dfrac{3}{1}.\dfrac{{\sqrt x  - 2}}{{3\sqrt x  + 3}}\\
 = \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}}\\
b)P = \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}} = \dfrac{{\sqrt x  + 1 - 3}}{{\sqrt x  + 1}}\\
 = 1 - \dfrac{3}{{\sqrt x  + 1}}\\
Do:\dfrac{3}{{\sqrt x  + 1}} > 0\\
 \Rightarrow 1 - \dfrac{3}{{\sqrt x  + 1}} < 1\\
 \Rightarrow P < 1\\
B3)\\
a)Dkxd:x > 0;x \ne 1\\
G = \left( {\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} + \dfrac{{\sqrt x }}{{\sqrt x  + 1}} + \dfrac{{\sqrt x }}{{1 - x}}} \right)\\
:\left( {\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} + \dfrac{{1 - \sqrt x }}{{\sqrt x  + 1}}} \right)\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2} + \sqrt x \left( {\sqrt x  - 1} \right) - \sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
:\dfrac{{{{\left( {\sqrt x  + 1} \right)}^2} - {{\left( {\sqrt x  - 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2\sqrt x  + 1 + x - \sqrt x  - \sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
.\dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{4\sqrt x }}\\
 = \dfrac{{2x + 1}}{{4\sqrt x }}\\
b)x = \dfrac{{2 - \sqrt 3 }}{2} = \dfrac{{4 - 2\sqrt 3 }}{4} = {\left( {\dfrac{{\sqrt 3  - 1}}{2}} \right)^2}\\
 \Rightarrow \sqrt x  = \dfrac{{\sqrt 3  - 1}}{2}\\
 \Rightarrow P = \dfrac{{2.\dfrac{{2 - \sqrt 3 }}{2} + 1}}{{4.\dfrac{{\sqrt 3  - 1}}{2}}} = \dfrac{{2 - \sqrt 3  + 1}}{{2\sqrt 3  - 2}}\\
 = \dfrac{{\sqrt 3 \left( {\sqrt 3  - 1} \right)}}{{2\left( {\sqrt 3  - 1} \right)}} = \dfrac{{\sqrt 3 }}{2}\\
c)G - \dfrac{1}{2}\\
 = \dfrac{{2x + 1}}{{4\sqrt x }} - \dfrac{1}{2}\\
 = \dfrac{{2x + 1 - 2\sqrt x }}{{4\sqrt x }}\\
 = \dfrac{{2\left( {x - \sqrt x  + \dfrac{1}{4}} \right) + \dfrac{1}{2}}}{{4\sqrt x }}\\
 = \dfrac{{2{{\left( {\sqrt x  - \dfrac{1}{2}} \right)}^2} + \dfrac{1}{2}}}{{4\sqrt x }} > 0\\
 \Rightarrow G > \dfrac{1}{2}\\
d)G = 0\\
 \Rightarrow \dfrac{{2x + 1}}{{4\sqrt x }} = 0 \Rightarrow x =  - \dfrac{1}{2}\left( {ktm} \right)
\end{array}$

Vậy ko có x để G=0

$\begin{array}{l}
4)a)Dkxd:x > 0\\
M = \dfrac{{x - 2}}{{x + 2\sqrt x }} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x  + 2}}\\
 = \dfrac{{x - 2 - \left( {\sqrt x  + 2} \right) + \sqrt x }}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 2 - \sqrt x  - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 4}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 2}}{{\sqrt x }}\\
b)M = \dfrac{{\sqrt x  - 2}}{{\sqrt x }} = 1 - \dfrac{2}{{\sqrt x }} < 1\\
 \Rightarrow M < 1
\end{array}$