Đáp án:
c. x=16
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
D = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 2\left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 6\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 4\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}}\\
b.D < 1\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} < 1\\
\to \dfrac{{\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}} < 0\\
\to \dfrac{1}{{\sqrt x - 3}} < 0\\
\to \sqrt x - 3 < 0\\
\to x < 9\\
\to 0 \le x < 9;x \ne 4\\
c.D = \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 1}}{{\sqrt x - 3}} = 1 + \dfrac{1}{{\sqrt x - 3}}\\
D \in Z\\
\to \dfrac{1}{{\sqrt x - 3}} \in Z\\
\to \sqrt x - 3 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 16\\
x = 4\left( l \right)
\end{array} \right.
\end{array}\)
