Đáp án:
c) Min=-1
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 9\\
P = \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}:\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= - \dfrac{3}{{\sqrt x + 3}}\\
b)P < - \dfrac{1}{2}\\
\to - \dfrac{3}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{3 - \sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 9 > x \ge 0\\
c)P = - \dfrac{3}{{\sqrt x + 3}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 3 \ge 3\\
\to \dfrac{3}{{\sqrt x + 3}} \le 1\\
\to - \dfrac{3}{{\sqrt x + 3}} \ge - 1\\
\to Min = - 1\\
\Leftrightarrow x = 0
\end{array}\)
