Đáp án:

\(\begin{array}{l}
B3:\\
a)A = \dfrac{{2 - 5\sqrt x }}{{\sqrt x  + 3}}\\
b)x = \dfrac{1}{{121}}\\
c)x \in \emptyset \\
B6:\\
a)M = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
b)x \in \emptyset \\
c)\left[ \begin{array}{l}
x = 9\\
x = 4
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B3:\\
a)DK:x \ge 0;x \ne 1\\
A = \dfrac{{15\sqrt x  - 11 - \left( {3\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right) - \left( {2\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{15\sqrt x  - 11 - 3x - 7\sqrt x  + 6 - 2x - \sqrt x  + 3}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{ - 5x + 7\sqrt x  - 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {2 - 5\sqrt x } \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2 - 5\sqrt x }}{{\sqrt x  + 3}}\\
b)A = \dfrac{1}{2}\\
 \to \dfrac{{2 - 5\sqrt x }}{{\sqrt x  + 3}} = \dfrac{1}{2}\\
 \to 4 - 10\sqrt x  = \sqrt x  + 3\\
 \to 11\sqrt x  = 1\\
 \to \sqrt x  = \dfrac{1}{{11}}\\
 \to x = \dfrac{1}{{121}}\\
c)A = \dfrac{{2 - 5\sqrt x }}{{\sqrt x  + 3}} =  - \dfrac{{5\left( {\sqrt x  + 3} \right) - 17}}{{\sqrt x  + 3}}\\
 =  - 5 + \dfrac{{17}}{{\sqrt x  + 3}}\\
A \in Z \to \dfrac{{17}}{{\sqrt x  + 3}} \in Z\\
 \to \sqrt x  + 3 \in U\left( {17} \right)\\
 \to \sqrt x  + 3 = 17\\
 \to \sqrt x  = 14\\
 \to x = 196 \to A =  - 5 + \dfrac{{17}}{{\sqrt {196}  + 3}} =  - 4\left( {KTM} \right)\\
 \to x \in \emptyset \\
B6:\\
a)M = \dfrac{{3x + 3\sqrt x  - 3 - \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{{\sqrt x  - 2}}{{\sqrt x }}.\dfrac{{1 - 1 + \sqrt x }}{{1 - \sqrt x }}\\
 = \dfrac{{3x + 3\sqrt x  - 3 - x + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}} - \dfrac{{\sqrt x  - 2}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x  - 1}}\\
 = \dfrac{{2x + 3\sqrt x  - 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}} - \dfrac{{\sqrt x  - 2}}{{\sqrt x  - 1}}\\
 = \dfrac{{2x + 3\sqrt x  - 2 - \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{2x + 3\sqrt x  - 2 - x + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x + 3\sqrt x  + 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
b)M = \dfrac{1}{2}\\
 \to \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} = \dfrac{1}{2}\\
 \to 2\sqrt x  + 2 = \sqrt x  - 1\\
 \to \sqrt x  =  - 3\left( {KTM} \right)\\
 \to x \in \emptyset \\
c)M = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} = \dfrac{{\sqrt x  - 1 + 2}}{{\sqrt x  - 1}}\\
 = 1 + \dfrac{2}{{\sqrt x  - 1}}\\
M \in N \Leftrightarrow \dfrac{2}{{\sqrt x  - 1}} \in N\\
 \Leftrightarrow \sqrt x  - 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  - 1 = 2\\
\sqrt x  - 1 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 3\\
\sqrt x  = 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = 4
\end{array} \right. \to \left[ \begin{array}{l}
M = 1 + \dfrac{2}{{\sqrt 9  - 1}} = 2\\
M = 1 + \dfrac{2}{{\sqrt 4  - 1}} = 3
\end{array} \right.\left( {TMDK} \right)\\
KL:\left[ \begin{array}{l}
x = 9\\
x = 4
\end{array} \right.
\end{array}\)