Chọn phương án A.

Đặt $A=\lim\limits_{x\to-\sqrt{3}}\dfrac{2x^3+6\sqrt{3}}{3-x^2}$, ta có $$A=\dfrac{a\sqrt{3}}{b}\Leftrightarrow\dfrac{A}{\sqrt{3}}=\dfrac{a}{b}.$$

1. Tính giới hạn $A=\lim\limits_{x\to-\sqrt{3}}\dfrac{2x^3+6\sqrt{3}}{3-x^2}$.
   
   
   
   
   Vậy $A\approx5.194652567$.
2. Khi đó $\dfrac{a}{b}=2.999\approx3=\dfrac{3}{1}$.

Vậy $a=3$ và $b=1$. Do đó $a^2+b^2=10$.

Chọn phương án A.

\(\begin{aligned}
\lim\limits_{x\to-\sqrt{3}}\dfrac{2x^3+6\sqrt{3}}{3-x^2}&=\lim\limits_{x\to-\sqrt{3}}\dfrac{2\left(x^3+3\sqrt{3}\right)}{\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)}\\
&=\lim\limits_{x\to-\sqrt{3}}\dfrac{2\left(x+\sqrt{3}\right)\left(x^2-\sqrt{3}x+3\right)}{\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)}\\
&=\lim\limits_{x\to-\sqrt{3}}\dfrac{2\left(x^2-\sqrt{3}x+3\right)}{\sqrt{3}-x}\\
&=\dfrac{2\left(\left(-\sqrt{3}\right)^2-\sqrt{3}\cdot\left(-\sqrt{3}\right)+3\right)}{\sqrt{3}-\left(-\sqrt{3}\right)}\\
&=3\sqrt{3}=\dfrac{3\sqrt{3}}{1}.
\end{aligned}\)

Theo đó ta có \(\begin{cases}
a=3\\ b=1.
\end{cases}\)

Khi đó \(a^2+b^2=3^2+1^2=10\).