Đáp án:
g. MinM=-3
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
M = \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b.Thay:x = \dfrac{2}{{3 - \sqrt 5 }} = \dfrac{{2\left( {3 + \sqrt 5 } \right)}}{{9 - 5}} = \dfrac{{6 + 2\sqrt 5 }}{4}\\
= \dfrac{{5 + 2.\sqrt 5 .1 + 1}}{4} = \dfrac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{4}\\
M = \dfrac{{\sqrt {\dfrac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{4}} + 1}}{{\sqrt {\dfrac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{4}} - 3}} = \dfrac{{\dfrac{{\sqrt 5 + 1 + 2}}{2}}}{{\dfrac{{\sqrt 5 + 1 - 6}}{2}}}\\
= \dfrac{{\sqrt 5 + 3}}{{\sqrt 5 - 5}}\\
c.M < 1\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} < 1\\
\to \dfrac{{\sqrt x + 1 - \sqrt x + 3}}{{\sqrt x - 3}} < 0\\
\to \dfrac{4}{{\sqrt x - 3}} < 0\\
\to \sqrt x - 3 < 0\\
\to 0 \le x < 9;x \ne 4\\
d.M \in Z\\
Ma:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}} = 1 + \dfrac{4}{{\sqrt x - 3}}\\
\Leftrightarrow \dfrac{4}{{\sqrt x - 3}} \in Z\\
\Leftrightarrow \sqrt x - 3 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 4\\
\sqrt x - 3 = - 4\left( l \right)\\
\sqrt x - 3 = 2\\
\sqrt x - 3 = - 2\\
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 7\\
\sqrt x = 5\\
\sqrt x = 1\\
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 49\\
x = 25\\
x = 1\\
x = 16\\
x = 4\left( l \right)
\end{array} \right.\\
e.M = 3\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = 3\\
\to \sqrt x + 1 = 3\sqrt x - 9\\
\to 2\sqrt x = 10\\
\to \sqrt x = 5\\
\to x = 25\\
f.M = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}} = 1 + \dfrac{4}{{\sqrt x - 3}}\\
Do:x \ge 0 \to \sqrt x \ge 0\\
\to \sqrt x - 3 \ge - 3\\
\to \dfrac{4}{{\sqrt x - 3}} \le - \dfrac{4}{3}\\
\to 1 + \dfrac{4}{{\sqrt x - 3}} \le 1 - \dfrac{4}{3} = - \dfrac{1}{3}\\
\to M < 1\\
g.\dfrac{1}{M} = \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 4}}{{\sqrt x + 1}} = 1 - \dfrac{4}{{\sqrt x + 1}}
\end{array}\)
Để M đạt GTNN
⇔\(\dfrac{4}{{\sqrt x + 1}}\) đạt GTLN
⇔ \({\sqrt x + 1}\) đạt GTNN
\(\begin{array}{l}
\to \sqrt x + 1 = 1\\
\to \sqrt x = 0\\
\to x = 0\\
\to MinM = 1 - \dfrac{4}{{\sqrt 0 + 1}} = - 3
\end{array}\)
