Đáp án:
c) $x \in \left\{ {0;1;16} \right\}$
Giải thích các bước giải:
a) ĐKXĐ: $x\ge 0;x\ne 4$
Ta có:
$\begin{array}{l}
A = \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} - \dfrac{5}{{x + \sqrt x - 6}} + \dfrac{1}{{2 - \sqrt x }}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} - \dfrac{5}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - 5 - \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 - 5 - \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 12}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}}\\
= 1 - \dfrac{2}{{\sqrt x - 2}}
\end{array}$
c) Ta có:
$\begin{array}{l}
A \in N\\
\Leftrightarrow \left\{ \begin{array}{l}
A \ge 0\\
\left( {\sqrt x - 2} \right) \in U\left( 2 \right)
\end{array} \right.\left( {Do:x \in Z} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - \dfrac{2}{{\sqrt x - 2}} \ge 0\\
\sqrt x - 2 \in \left\{ { - 2; - 1;1;2} \right\}
\end{array} \right.\\
\Leftrightarrow \left( {\sqrt x - 2} \right) \in \left\{ { - 2; - 1;2} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;1;4} \right\}\\
\Leftrightarrow x \in \left\{ {0;1;16} \right\}(tm)
\end{array}$
Vậy $x \in \left\{ {0;1;16} \right\}$
