Đáp án:
\(\begin{array}{l}
a)x \ge 0;x \ne \left\{ {4;9} \right\}\\
b)\dfrac{3}{{\sqrt x - 2}}\\
c)\left[ \begin{array}{l}
0 \le x < 4\\
x > 5;x \ne 9
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = 25\\
x = 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
b)A = \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} - 1} \right]:\dfrac{{9 - x + \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) - {{\left( {\sqrt x - 2} \right)}^2}}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - \sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}{{9 - x + x - 9 - {{\left( {\sqrt x - 2} \right)}^2}}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}.\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}{{ - {{\left( {\sqrt x - 2} \right)}^2}}}\\
= \dfrac{3}{{\sqrt x - 2}}\\
c)A < 1\\
\to \dfrac{3}{{\sqrt x - 2}} < 1\\
\to \dfrac{{3 - \sqrt x + 2}}{{\sqrt x - 2}} < 0\\
\to \dfrac{{5 - \sqrt x }}{{\sqrt x - 2}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
5 - \sqrt x > 0\\
\sqrt x - 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
5 - \sqrt x < 0\\
\sqrt x - 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x < 2\\
\sqrt x > 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
0 \le x < 4\\
x > 5;x \ne 9
\end{array} \right.\\
d)A = \dfrac{3}{{\sqrt x - 2}}\\
A \in Z \to \dfrac{3}{{\sqrt x - 2}} \in Z\\
\to \sqrt x - 2 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 2 = 3\\
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1\\
\sqrt x - 2 = - 3\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 25\\
x = 9\left( l \right)\\
x = 1
\end{array} \right.
\end{array}\)
