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