Chọn phương án D.

\(\begin{aligned}
&\lim\limits_{x\to+\infty}x\left(\sqrt{4x^2+7x}+2x\right)\\
=&\lim\limits_{x\to+\infty}x\left(\sqrt{x^2\left(4+\dfrac{7}{x}\right)}+2x\right)\\
=&\lim\limits_{x\to+\infty}x\left(|x|\sqrt{4+\dfrac{7}{x}}+2x\right)\\
=&\lim\limits_{x\to+\infty}x\left(x\sqrt{4+\dfrac{7}{x}}+2x\right)\\
=&\lim\limits_{x\to+\infty}x^2\left(\sqrt{4+\dfrac{7}{x}}+2\right)\\
=&+\infty.
\end{aligned}\)

Vì \(\begin{cases}
\lim\limits_{x\to+\infty}x^2&=+\infty\\
\lim\limits_{x\to+\infty}\left(\sqrt{4+\dfrac{7}{x}}+2\right)&=4>0.
\end{cases}\)