Đáp án:
a. \(\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Q = \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b.Q = 2\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = 2\\
\to \sqrt x + 1 = 2\sqrt x - 6\\
\to \sqrt x = 7\\
\to x = 49\\
c.Q = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}} = 1 + \dfrac{4}{{\sqrt x - 3}}\\
Q \in Z \Leftrightarrow \dfrac{4}{{\sqrt x - 3}} \in Z\\
\Leftrightarrow \sqrt x - 3 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 4\\
\sqrt x - 3 = - 4\left( l \right)\\
\sqrt x - 3 = 2\\
\sqrt x - 3 = - 2\\
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 7\\
\sqrt x = 5\\
\sqrt x = 1\\
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 49\\
x = 25\\
x = 1\\
x = 16\\
x = 4\left( l \right)
\end{array} \right.
\end{array}\)
