Đáp án:
3) \(0 \le x < 4\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0\\
\left( {2\sqrt x - 1} \right)\left( {3\sqrt x - 6} \right) = 0\\
\to \left[ \begin{array}{l}
2\sqrt x - 1 = 0\\
3\sqrt x - 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 4
\end{array} \right.\\
2)DK:x \ge 0;x \ne 9\\
\dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt x - 2 > 0\\
\sqrt x - 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt x - 2 < 0\\
\sqrt x - 3 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 4\\
x > 9
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 4\\
x < 9
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 9\\
0 \le x < 4
\end{array} \right.\\
3)DK:x \ge 0;x \ne 4\\
\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 4
\end{array}\)
