Đáp án:
B3:
c. \(x > \dfrac{3}{4};x \ne \left\{ {4;9} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a.DK:x > 0;x \ne \left\{ {4;9} \right\}\\
b.Q = \left[ {\dfrac{{\sqrt x - \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right]:\left[ {\dfrac{{x - 9 - x + 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}} \right]\\
= \dfrac{3}{{\sqrt x \left( {\sqrt x - 3} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{ - 5}}\\
= - \dfrac{{3\left( {\sqrt x - 2} \right)}}{{5\sqrt x }}\\
c.Q < 1\\
\to - \dfrac{{3\left( {\sqrt x - 2} \right)}}{{5\sqrt x }} < 1\\
\to \dfrac{{ - 3\sqrt x + 6 - 5\sqrt x }}{{5\sqrt x }} < 0\\
\to - 8\sqrt x + 6 < 0\left( {do:5\sqrt x > 0\forall x > 0} \right)\\
\to 8\sqrt x > 6\\
\to x > \dfrac{3}{4};x \ne \left\{ {4;9} \right\}
\end{array}\)
