Đáp án:
a) \(\dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
B = \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {2 - 5\sqrt x } \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)B = \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{{ - 5\left( {\sqrt x + 3} \right) + 17}}{{\sqrt x + 3}}\\
= - 5 + \dfrac{{17}}{{\sqrt x + 3}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 3 \ge 3\\
\to \dfrac{{17}}{{\sqrt x + 3}} \le \dfrac{{17}}{3}\\
\to - 5 + \dfrac{{17}}{{\sqrt x + 3}} \le \dfrac{2}{3}\\
\to Max = \dfrac{2}{3}\\
\Leftrightarrow x = 0
\end{array}\)
⇒ Không có GTNN
