24. It is given that $x=\frac{1}{2-\sqrt{3}}$. Find the value of
$$
x^{6}-2 \sqrt{3} x^{5}-x^{4}+x^{3}-4 x^{2}+2 x-\sqrt{3} \text {. }
$$
24. Ans: 2
$$
x=\frac{1}{2-\sqrt{3}}=\frac{1}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3} .
$$

Then $x-2=\sqrt{3} \Longrightarrow x-\sqrt{3}=2 \longrightarrow x^{2}-4 x+1=0$ and $x^{2}-2 \sqrt{3} x-1=0$.
Thus,
$$
\begin{aligned}
& x^{6}-2 \sqrt{3} x^{5}-x^{4}+x^{3}-4 x^{2}+2 x-\sqrt{3} \\
= & x^{4}\left(x^{2}-2 \sqrt{3} x-1\right)+x\left(x^{2}-4 x+1\right)+x-\sqrt{3} \\
= & 0+0+2+\sqrt{3}-\sqrt{3}=2 .
\end{aligned}
$$