Đáp án:
c. \(\left[ \begin{array}{l}
x = 9\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0\\
M = \left( {\dfrac{{2x + 3\sqrt x }}{{x\sqrt x + 1}} + \dfrac{1}{{x - \sqrt x + 1}} - \dfrac{1}{{\sqrt x + 1}}} \right).\left( {\dfrac{{x - \sqrt x + 1}}{{\sqrt x }}} \right)\\
= \left[ {\dfrac{{2x + 3\sqrt x + \sqrt x + 1 - x + \sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right].\dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{x + 5\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\sqrt x + 5}}{{\sqrt x + 1}}\\
b.M = \dfrac{{\sqrt x + 5}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 + 4}}{{\sqrt x + 1}} = 1 + \dfrac{4}{{\sqrt x + 1}}\\
Do:x > 0 \to \sqrt x > 0\\
\to \sqrt x + 1 > 1\\
\to 0 < \dfrac{4}{{\sqrt x + 1}} < 4\\
\to 1 < 1 + \dfrac{4}{{\sqrt x + 1}} < 5\\
\to M > 1\\
c.Để:M \in Z\\
\Leftrightarrow \dfrac{4}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 4\\
\sqrt x + 1 = 2\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 1\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
