Đáp án:
b. \(\dfrac{{\sqrt x }}{{\sqrt x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
b.Q = \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x }}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 2\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 4\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
c.Q = \dfrac{{\sqrt x }}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 3}}{{\sqrt x - 3}} = 1 + \dfrac{3}{{\sqrt x - 3}}\\
Q \in Z \Leftrightarrow \dfrac{3}{{\sqrt x - 3}} \in Z\\
\Leftrightarrow \sqrt x - 3 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 3\\
\sqrt x - 3 = - 3\\
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 6\\
\sqrt x = 0\\
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 36\\
x = 0\\
x = 16\\
x = 4\left( l \right)
\end{array} \right.
\end{array}\)
