Đáp án:
d. Min=-1
Giải thích các bước giải:
\(\begin{array}{l}
a.C = \left[ {\dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}} \right)\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b.Thay:x = \dfrac{1}{4}\\
\to C = \dfrac{{ - 3}}{{\sqrt {\dfrac{1}{4}} + 3}} = \dfrac{{ - 6}}{7}\\
c.C < - \dfrac{1}{3}\\
\to \dfrac{{ - 3}}{{\sqrt x + 3}} < - \dfrac{1}{3}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{3}\\
\to \dfrac{{9 - \sqrt x - 3}}{{3\left( {\sqrt x + 3} \right)}} > 0\\
\to 6 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 36 > x \ge 0;x \ne 9\\
d.Do:x \ge 0\\
\to \sqrt x \ge 0 \to \sqrt x + 3 \ge 3\\
\to \dfrac{3}{{\sqrt x + 3}} \le 1\\
\to - \dfrac{3}{{\sqrt x + 3}} \ge - 1\\
\to Min = - 1\\
\Leftrightarrow x = 0
\end{array}\)
