Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
ĐK:x \ge 0;x \ne 1\\
\frac{B}{A} = \frac{{3\sqrt x }}{{\sqrt x + 2}}:\frac{{\sqrt x }}{{\sqrt x - 1}} = \frac{{3\sqrt x }}{{\sqrt x + 2}}.\frac{{\sqrt x - 1}}{{\sqrt x }} = \frac{{3(\sqrt x - 1)}}{{\sqrt x + 2}}\\
= \frac{{3\left( {\sqrt x + 2 - 3} \right)}}{{\sqrt x + 2}} = 3 - \frac{9}{{\sqrt x + 2}}
\end{array}\)
Để B/A nguyên
\(\begin{array}{l}
\Leftrightarrow \sqrt x + 2 \in U\left( 9 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 9\\
\sqrt x + 2 = - 9\left( l \right)\\
\sqrt x + 2 = 3\\
\sqrt x + 2 = - 3\left( l \right)\\
\sqrt x + 2 = 1\left( l \right)\\
\sqrt x + 2 = - 1\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 7\\
\sqrt x = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 49\left( {TM} \right)\\
x = 1\left( l \right)
\end{array} \right.
\end{array}\)
