\[ \lim_{x \to -\infty} \left( \sqrt{4x^2 + 3x + 1 + mx} \right) = +\infty \text{ nếu } \sqrt{4x^2 + 3x + 1 - mx} \]

To analyze the limit

\[
\lim_{x \to -\infty} \left( \sqrt{4x^2 + 3x + 1 + mx} \right),
\]

we start by factoring out \( x^2 \) from the expression inside the square root:

\[
\sqrt{4x^2 + 3x + 1 + mx} = \sqrt{x^2 \left( 4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x} \right)} = |x| \sqrt{4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x}}.
\]

As \( x \to -\infty \), \( |x| = -x \), so we have:

\[
-x \sqrt{4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x}}.
\]

Next, we analyze the limit of the square root as \( x \to -\infty \):

\[
\sqrt{4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x}} \to \sqrt{4} = 2.
\]

Thus, the limit becomes:

\[
\lim_{x \to -\infty} -x \cdot 2 = +\infty,
\]

if \( \sqrt{4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x}} > 0\).

Now, we need to analyze the condition under which

\[
\sqrt{4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x}} = 0.
\]

This occurs when:

\[
4 + \frac{3}{x} + \frac{1}{x^2} + \frac{m}{x} \leq 0.
\]

As \( x \to -\infty \), the terms \( \frac{3}{x}, \frac{1}{x^2}, \frac{m}{x} \) become negligible, leading to:

\[
4 + \text{(small negative terms)} > 0.
\]

Thus, \( 4 > 0 \) will always hold. Therefore, we focus on when the inner square root is still positive:

From the expression:

\[
4 + m < 0 \implies m < -4,
\]

is not the only concern since we need to ensure that when combined with other terms, it does not become non-positive. The additional conditions from other terms need careful checking, leading us to enforce that:

To ensure that the limit approaches \( +\infty \), we combine the conditions \( m > 2 \):

Thus, the correct choice is:

**B. \( m > 2 \)**.