Đáp án:

\(\begin{array}{l}
B1:\\
c)0 \le x < 4\\
d)\left[ \begin{array}{l}
x = 16\\
x = 0\\
x = 9\\
x = 1
\end{array} \right.\\
B2:\\
a)\dfrac{3}{{x + \sqrt x  + 1}}\\
b)x = \dfrac{{2 - \sqrt 3 }}{2}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
Q = \dfrac{{\sqrt x  - 4 + 2\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{\sqrt x  - 4 + 2\sqrt x  + 4}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{3\sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\
M = Q:P = \dfrac{{3\sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}:\dfrac{3}{{\sqrt x  + 2}}\\
 = \dfrac{{3\sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}.\dfrac{{\sqrt x  + 2}}{3}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\
c)M < \dfrac{1}{2}\\
 \to \dfrac{{\sqrt x }}{{\sqrt x  - 2}} < \dfrac{1}{2}\\
 \to \dfrac{{2\sqrt x  - \sqrt x  + 2}}{{2\left( {\sqrt x  - 2} \right)}} < 0\\
 \to \dfrac{{\sqrt x  + 2}}{{2\left( {\sqrt x  - 2} \right)}} < 0\\
 \to \sqrt x  - 2 < 0\left( {do:\sqrt x  + 2 > 0\forall x \ge 0} \right)\\
 \to 0 \le x < 4\\
d)M = \dfrac{{\sqrt x }}{{\sqrt x  - 2}} = \dfrac{{\sqrt x  - 2 + 2}}{{\sqrt x  - 2}}\\
 = 1 + \dfrac{2}{{\sqrt x  - 2}}\\
M \in Z \to \dfrac{2}{{\sqrt x  - 2}} \in Z\\
 \to \sqrt x  - 2 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  - 2 = 2\\
\sqrt x  - 2 =  - 2\\
\sqrt x  - 2 = 1\\
\sqrt x  - 2 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x  = 4\\
\sqrt x  = 0\\
\sqrt x  = 3\\
\sqrt x  = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 16\\
x = 0\\
x = 9\\
x = 1
\end{array} \right.\\
B2:\\
a)M = \dfrac{{x + \sqrt x  + 1 - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{3\left( {\sqrt x  - 1} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{3\left( {\sqrt x  - 1} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{3}{{x + \sqrt x  + 1}}\\
b)Do:\dfrac{3}{{x + \sqrt x  + 1}} > 0\forall x > 0\\
\dfrac{3}{{x + \sqrt x  + 1}} = \dfrac{3}{{x + 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}}}\\
 = \dfrac{3}{{{{\left( {\sqrt x  + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}\\
Do:{\left( {\sqrt x  + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge 0\forall x > 0\\
 \to \dfrac{3}{{{{\left( {\sqrt x  + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} < 4\\
 \to 0 < M < 4\\
Do:M \in Z \to \left[ \begin{array}{l}
M = 1\\
M = 2\\
M = 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{3}{{x + \sqrt x  + 1}} = 1\\
\dfrac{3}{{x + \sqrt x  + 1}} = 2\\
\dfrac{3}{{x + \sqrt x  + 1}} = 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x + \sqrt x  + 1 = 3\\
x + \sqrt x  + 1 = \dfrac{3}{2}\\
x + \sqrt x  + 1 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x + \sqrt x  - 2 = 0\\
x + \sqrt x  - \dfrac{1}{2} = 0\\
x + \sqrt x  = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right) = 0\\
\sqrt x  = \dfrac{{ - 1 + \sqrt 3 }}{2}\\
\sqrt x  = \dfrac{{ - 1 - \sqrt 3 }}{2}\left( l \right)\\
\sqrt x \left( {\sqrt x  + 1} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  - 1 = 0\\
\sqrt x  = \dfrac{{ - 1 + \sqrt 3 }}{2}\\
x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = \dfrac{{2 - \sqrt 3 }}{2}\\
x = 0\left( l \right)
\end{array} \right.\\
 \to x = \dfrac{{2 - \sqrt 3 }}{2}
\end{array}\)