Đáp án:
a. \(\frac{{\sqrt x - 2}}{{3 + \sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.M = \frac{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right) + {{\left( {\sqrt x - 2} \right)}^2} - 9 + x}}{{\left( {\sqrt x - 2} \right)\left( {3 + \sqrt x } \right)}}\\
= \frac{{9 - x + x - 4\sqrt x + 4 - 9 + x}}{{\left( {\sqrt x - 2} \right)\left( {3 + \sqrt x } \right)}}\\
= \frac{{x - 4\sqrt x + 4}}{{\left( {\sqrt x - 2} \right)\left( {3 + \sqrt x } \right)}}\\
= \frac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {3 + \sqrt x } \right)}}\\
= \frac{{\sqrt x - 2}}{{3 + \sqrt x }}\\
b.M = \frac{7}{{12}}\\
\to \frac{{\sqrt x - 2}}{{3 + \sqrt x }} = \frac{7}{{12}}\\
\to 12\sqrt x - 24 = 21 + 7\sqrt x \\
\to 5\sqrt x = 45\\
\to \sqrt x = 9\\
\to x = 81\left( {TM} \right)\\
c.M > \frac{1}{2}\\
\to \frac{{\sqrt x - 2}}{{3 + \sqrt x }} > \frac{1}{2}\\
\to \frac{{2\sqrt x - 4 - 3 - \sqrt x }}{{3 + \sqrt x }} > 0\\
\to \frac{{\sqrt x - 7}}{{3 + \sqrt x }} > 0\\
\to \sqrt x - 7 > 0\left( {do:3 + \sqrt x > 0\forall x \ge 0} \right)\\
\to x > 49\\
d.\frac{1}{M} = \frac{{\sqrt x + 3}}{{\sqrt x - 2}} = \frac{{\sqrt x - 2 + 5}}{{\sqrt x - 2}} = 1 + \frac{5}{{\sqrt x - 2}}\\
Để:\frac{1}{M} \in Z\\
\to \frac{5}{{\sqrt x - 2}} \in Z\\
\to \sqrt x - 2 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 2 = 5\\
\sqrt x - 2 = - 5\left( l \right)\\
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 49\\
x = 9\\
x = 1
\end{array} \right.\\
e.M = \frac{{\sqrt x - 2}}{{3 + \sqrt x }} = \frac{{3 + \sqrt x - 5}}{{3 + \sqrt x }} = 1 - \frac{5}{{3 + \sqrt x }}\\
Để:M \in Z\\
\to \frac{5}{{3 + \sqrt x }} \in Z\\
\to 3 + \sqrt x \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
3 + \sqrt x = 5\\
3 + \sqrt x = 1\left( l \right)
\end{array} \right.\left( {do:3 + \sqrt x > 0\forall x \ge 0} \right)\\
\to x = 4\left( l \right)
\end{array}\)
⇒ Không tồn tại x để M nguyên
