\(\sqrt{x^2-3}+3\le x^2\Rightarrow\sqrt{x^2-3}\le x^2-3\Rightarrow1\le\dfrac{x^2-3}{\sqrt{x^2-3}}=\left(....\right)\Rightarrow1\le\sqrt{x^2-3}\Rightarrow1\le x^2-3\Rightarrow x^2\ge4\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)