Đáp án:
e) \(Min = - \dfrac{{1003}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0\\
P = x - 2.\sqrt x .\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{{65}}{4}\\
= {\left( {\sqrt x - \dfrac{3}{2}} \right)^2} - \dfrac{{65}}{4}\\
Do:{\left( {\sqrt x - \dfrac{3}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x - \dfrac{3}{2}} \right)^2} - \dfrac{{65}}{4} \ge - \dfrac{{65}}{4}\\
\to Min = - \dfrac{{65}}{4}\\
\Leftrightarrow \sqrt x - \dfrac{3}{2} = 0\\
\Leftrightarrow x = \dfrac{9}{4}\\
c)DK:x \ge 0\\
B = \dfrac{{2\left( {x - 2\sqrt x - 4} \right) - 5}}{{ - \left( {x - 2\sqrt x - 4} \right)}}\\
= - 2 + \dfrac{5}{{x - 2\sqrt x - 4}}\\
= - 2 + \dfrac{5}{{x - 2\sqrt x + 1 - 5}}\\
= - 2 + \dfrac{5}{{{{\left( {\sqrt x - 1} \right)}^2} - 5}}\\
Do:{\left( {\sqrt x - 1} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x - 1} \right)^2} - 5 \ge - 5\\
\to \dfrac{5}{{{{\left( {\sqrt x - 1} \right)}^2} - 5}} \le - 1\\
\to - 2 + \dfrac{5}{{{{\left( {\sqrt x - 1} \right)}^2} - 5}} \le - 3\\
\to Max = - 3\\
\Leftrightarrow \sqrt x - 1 = 0\\
\Leftrightarrow x = 1\\
d)DK:x \ge 0\\
P = - \dfrac{3}{{\sqrt x + 2}}\\
Do:x \ge 0\\
\to \sqrt x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{3}{{\sqrt x + 2}} \le \dfrac{3}{2}\\
\to - \dfrac{3}{{\sqrt x + 2}} \ge - \dfrac{3}{2}\\
\to Min = - \dfrac{3}{2}\\
\Leftrightarrow x = 0\\
e)P = - \dfrac{{2006}}{{\sqrt x + 8}}\\
Do:x \ge 0\\
\to \sqrt x \ge 0\\
\to \sqrt x + 8 \ge 8\\
\to \dfrac{{2006}}{{\sqrt x + 8}} \le \dfrac{{2006}}{8}\\
\to - \dfrac{{2006}}{{\sqrt x + 8}} \ge - \dfrac{{1003}}{4}\\
\to Min = - \dfrac{{1003}}{4}\\
\Leftrightarrow x = 0
\end{array}\)
\(\begin{array}{l}
b)A = \dfrac{1}{{2x - 2.\sqrt {2x} .\dfrac{3}{{\sqrt 2 }} + \dfrac{9}{2} + \dfrac{{17}}{2}}}\\
= \dfrac{1}{{{{\left( {\sqrt {2x} {\rm{\;}} - \dfrac{3}{{\sqrt 2 }}} \right)}^2} + \dfrac{{17}}{2}}}\\
Do:{\left( {\sqrt {2x} {\rm{\;}} - \dfrac{3}{{\sqrt 2 }}} \right)^2} \ge 0\forall x \ge 0{\rm{\;}}\\
\to {\left( {\sqrt {2x} {\rm{\;}} - \dfrac{3}{{\sqrt 2 }}} \right)^2} + \dfrac{{17}}{2} \ge \dfrac{{17}}{2}\\
\to \dfrac{1}{{{{\left( {\sqrt {2x} {\rm{\;}} - \dfrac{3}{{\sqrt 2 }}} \right)}^2} + \dfrac{{17}}{2}}} \le \dfrac{2}{{17}}\\
\to Max = \dfrac{2}{{17}}\\
\Leftrightarrow \sqrt {2x} {\rm{\;}} - \dfrac{3}{{\sqrt 2 }} = 0\\
\to \sqrt x = \dfrac{3}{2}\\
\to x = \dfrac{9}{4}
\end{array}\)
