Đáp án:

B5:

b) \(0 \le x < 9\)

Giải thích các bước giải:

\(\begin{array}{l}
B3:\\
a)DK:x > 0\\
A = \left[ {\dfrac{{2\sqrt x  + x + 1}}{{\sqrt x }}} \right].\left[ {\dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  - 1} \right)}}} \right]\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}.\dfrac{{x + \sqrt x  - 2 - x + \sqrt x  + 2}}{{{{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{2\sqrt x }}{{\sqrt x \left( {\sqrt x  - 1} \right)}} = \dfrac{2}{{\sqrt x  - 1}}\\
b)Thay:x = 2\\
 \to A = \dfrac{2}{{\sqrt 2  - 1}} = 2\left( {\sqrt 2  + 1} \right) = 2 + 2\sqrt 2 \\
c)A < 1\\
 \to \dfrac{2}{{\sqrt x  - 1}} < 1\\
 \to \dfrac{{2 - \sqrt x  + 1}}{{\sqrt x  - 1}} < 0\\
 \to \dfrac{{3 - \sqrt x }}{{\sqrt x  - 1}} < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 - \sqrt x  > 0\\
\sqrt x  - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3 - \sqrt x  < 0\\
\sqrt x  - 1 > 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 > \sqrt x \\
\sqrt x  < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3 < \sqrt x \\
\sqrt x  > 1
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x < 1\\
x > 9
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
B5:\\
a)DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
Q = \dfrac{{2\sqrt x  - 9 - \left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) + \left( {2\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{2\sqrt x  - 9 - x + 9 + 2x - 3\sqrt x  - 2}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{x - \sqrt x  - 2}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\
b)Q < 1\\
 \to \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} < 1\\
 \to \dfrac{{\sqrt x  + 1 - \sqrt x  + 3}}{{\sqrt x  - 3}} < 0\\
 \to \dfrac{4}{{\sqrt x  - 3}} < 0\\
 \to \sqrt x  - 3 < 0\\
 \to x < 9\\
 \to 0 \le x < 9\\
c)Q = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} = \dfrac{{\sqrt x  - 3 + 4}}{{\sqrt x  - 3}} = 1 + \dfrac{4}{{\sqrt x  - 3}}\\
Q \in Z \Leftrightarrow \dfrac{4}{{\sqrt x  - 3}} \in Z\\
 \Leftrightarrow \sqrt x  - 3 \in U\left( 4 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  - 3 = 4\\
\sqrt x  - 3 =  - 4\left( l \right)\\
\sqrt x  - 3 = 2\\
\sqrt x  - 3 =  - 2\\
\sqrt x  - 3 = 1\\
\sqrt x  - 3 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 49\\
x = 25\\
x = 1\\
x = 16\\
x = 4\left( l \right)
\end{array} \right.
\end{array}\)