Tìm x để $\frac{8√x+8}{6√x+9}$ > $\frac{8}{3}$

Đáp án:

Giải thích các bước giải:

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