Đáp án:
\(x \ge 0\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
P = - \dfrac{3}{{\sqrt x + 3}}\\
P < \dfrac{1}{2}\\
\to - \dfrac{3}{{\sqrt x + 3}} < \dfrac{1}{2}\\
\to \dfrac{{ - 6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} < 0\\
\to \dfrac{{ - 9 - \sqrt x }}{{2\left( {\sqrt x + 3} \right)}} < 0\\
\to - \dfrac{{\sqrt x + 9}}{{2\left( {\sqrt x + 3} \right)}} < 0\\
\to \dfrac{{\sqrt x + 9}}{{2\left( {\sqrt x + 3} \right)}} > 0\left( {ld} \right)\\
Do:\left\{ \begin{array}{l}
\sqrt x + 9 > 0\forall x \ge 0\\
\sqrt x + 3 > 0\forall x \ge 0
\end{array} \right.\\
KL:x \ge 0
\end{array}\)
