Giải thích các bước giải:
\(\begin{array}{l}
y = \frac{{\sqrt {m{x^2} + 3mx + 1} }}{{x + 2}}\\
Dk:\left\{ \begin{array}{l}
m{x^2} + 3mx + 1 \ge 0\\
x \ne - 2
\end{array} \right.\\
De\;ham\;so\;co\;3\;tiem\;can\;thi\;phai\;ton\;tai\mathop {\lim }\limits_{x \to - \infty } y \ne \mathop {\lim }\limits_{x \to + \infty } y;\mathop {\lim }\limits_{x \to 2} y\\
\Rightarrow Txd\;co\;dang:D = \left( { - \infty ;a} \right] \cup \left[ {b; + \infty } \right)\;va\;2 \in \left( { - \infty ;a} \right) \cup \left( {b; + \infty } \right)\\
m{.2^2} + 3m.2 + 1 \ne 0 \Leftrightarrow m \ne \frac{{ - 1}}{{10}}\\
TH1:Pt\,m{x^2} + 3mx + 1 = 0\;vo\;nghiem\;\\
\Rightarrow \Delta = 9{m^2} - 4m \le 0 \Leftrightarrow 0 < m < \frac{4}{9}\\
\to Ham\;so\;luon\;co\;3\;tiem\;can\\
Th2:Pt\;m{x^2} + 3mx + 1 = 0\;co\;2\;nghiem\;a,b\\
\Rightarrow \;\left\{ \begin{array}{l}
m > 0\\
\Delta > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > \frac{4}{9}\\
m < 0
\end{array} \right.\\
m > 0
\end{array} \right. \Rightarrow m > \frac{4}{9}.Theo\;vi - et\left\{ \begin{array}{l}
a + b = - 3\\
a.b = \frac{1}{m}
\end{array} \right.\\
De\;\exists \mathop {\lim }\limits_{x \to 2} y \Rightarrow \left[ \begin{array}{l}
2 < a < b\\
a < b < 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
a - 2 + b - 2 = - 3 - 4 > 0 \to loai\\
\left\{ \begin{array}{l}
a + b - 4 < 0\\
(a - 2)(b - 2) > 0
\end{array} \right. \Leftrightarrow \frac{1}{m} - 2.( - 3) + 4 > 0 \Leftrightarrow \frac{1}{m} > - 10 \to tm\forall m > 0
\end{array} \right.\\
Th3:Pt\;nghiem\;kep\\
m = \frac{4}{9} \Rightarrow x = \frac{{ - 3}}{2} \to tm\\
Vay\;m > 0\\
\end{array}\)
