Đáp án:
Giải thích các bước giải:
ĐKXĐ : x>0
\[\begin{array}{l}
\frac{{8\sqrt x + 8}}{{6\sqrt x + 9}} > \frac{8}{3}\\
\Leftrightarrow \frac{{8\sqrt x + 8}}{{6\sqrt x + 9}} - \frac{8}{3} > 0\\
\Leftrightarrow \frac{{8\sqrt x + 8 - 8\left( {2\sqrt x + 3} \right)}}{{3\left( {2\sqrt x + 3} \right)}} > 0\\
\Leftrightarrow \frac{{8\sqrt x + 8 - 16\sqrt x - 24}}{{3\left( {2\sqrt x + 3} \right)}} > 0\\
\Leftrightarrow \frac{{ - 8\sqrt x - 16}}{{3\left( {2\sqrt x + 3} \right)}} > 0\\
\Leftrightarrow \frac{{ - 8\left( {\sqrt x - 2} \right)}}{{3\left( {2\sqrt x + 3} \right)}} > 0\\
Vi\,\,\,3\left( {2\sqrt x + 3} \right) > 0\,\,\,\,\forall x \in D\\
nen\, - 8\left( {\sqrt x - 2} \right) > 0 \Leftrightarrow \sqrt x - 2 < 0 \Leftrightarrow \sqrt x < 2\\
Vay\,\,\,0 < x < 4
\end{array}\]
