Đáp án:
\(\begin{array}{l}
a)\dfrac{2}{{\sqrt x - 3}}\\
b)25 > x > 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 9\\
B = \left( {\dfrac{1}{{3 - \sqrt x }} - \dfrac{1}{{3 + \sqrt x }}} \right).\dfrac{{\sqrt x + 3}}{{\sqrt x }}\\
= \dfrac{{3 + \sqrt x - 3 + \sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{\sqrt x }}\\
= \dfrac{2}{{\sqrt x - 3}}\\
b)B > 1\\
\to \dfrac{2}{{\sqrt x - 3}} > 1\\
\to \dfrac{{2 - \sqrt x + 3}}{{\sqrt x - 3}} > 0\\
\to \dfrac{{5 - \sqrt x }}{{\sqrt x - 3}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
5 - \sqrt x > 0\\
\sqrt x - 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
5 - \sqrt x < 0\\
\sqrt x - 3 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
5 > \sqrt x > 3\\
\left\{ \begin{array}{l}
5 < \sqrt x \\
\sqrt x < 3
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 25 > x > 9
\end{array}\)
