Đáp án:
4) \(MinM = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 4\\
M = \dfrac{{x + 12 + \sqrt x - 2 - 4\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
3)\dfrac{1}{M} = \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 3}}{{\sqrt x - 1}}\\
= 1 + \dfrac{3}{{\sqrt x - 1}}\\
\dfrac{1}{M} \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \sqrt x - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 3\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1\\
\sqrt x - 1 = - 3\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 16\left( {TM} \right)\\
x = 4\left( l \right)\\
x = 0\left( {TM} \right)
\end{array} \right.\\
4)M = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 3}}{{\sqrt x + 2}}\\
= 1 - \dfrac{3}{{\sqrt x + 2}}
\end{array}\)
Để M đạt GTNN
⇔ \(\dfrac{3}{{\sqrt x + 2}}\) đạt GTLN
⇔ \({\sqrt x + 2}\) đạt GTNN
Mà:
\(\begin{array}{l}
\sqrt x + 2 \ge 2\\
\to \left( {\sqrt x + 2} \right)Min = 2\\
\Leftrightarrow x = 0\\
\to MinM = 1 - \dfrac{3}{2} = - \dfrac{1}{2}
\end{array}\)
