Đáp án:
d) Max=0
Giải thích các bước giải:
\(\begin{array}{l}
K = \dfrac{{15\sqrt x - 3}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x }}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
a)DK:x \ge 0;x \ne 1\\
b)K = \dfrac{{15\sqrt x - 3 - 3\sqrt x \left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 3 - 3x - 9\sqrt x - 2x - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 5\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}} = \dfrac{{ - 5\sqrt x }}{{\sqrt x + 3}}\\
c)Thay:K = \dfrac{1}{2}\\
\to \dfrac{{ - 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{2}\\
\to - 10\sqrt x = \sqrt x + 3\\
\to 11\sqrt x = - 3\left( l \right)\\
\to x \in \emptyset \\
d)K = \dfrac{{ - 5\sqrt x }}{{\sqrt x + 3}} = - \dfrac{{5\left( {\sqrt x + 3} \right) - 15}}{{\sqrt x + 3}}\\
= - 5 + \dfrac{{15}}{{\sqrt x + 3}}\\
Do:x \ge 0 \to \sqrt x \ge 0\\
\to \sqrt x + 3 \ge 3\\
\to \dfrac{{15}}{{\sqrt x + 3}} \le 5\\
\to - 5 + \dfrac{{15}}{{\sqrt x + 3}} \le 0\\
\to Max = 0\\
\Leftrightarrow x = 0
\end{array}\)
