Đáp án:
\(\dfrac{{19\sqrt x }}{{4\left( {\sqrt x + 4} \right)}} \ge 0\left( {ld} \right)\forall x \ge 0\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{ - 3\sqrt x + 7}}{{\sqrt x + 4}} \le \dfrac{7}{4}\\
\to \dfrac{{ - 12\sqrt x + 28 - 7\sqrt x - 28}}{{4\left( {\sqrt x + 4} \right)}} \le 0\\
\to \dfrac{{ - 19\sqrt x }}{{4\left( {\sqrt x + 4} \right)}} \le 0\\
\to \dfrac{{19\sqrt x }}{{4\left( {\sqrt x + 4} \right)}} \ge 0\left( {ld} \right)\forall x \ge 0\\
Do:x \ge 0 \to \sqrt x \ge 0\\
\to \left\{ \begin{array}{l}
19\sqrt x \ge 0\\
4\left( {\sqrt x + 4} \right) > 0
\end{array} \right.\\
\to dpcm
\end{array}\)
