find the value of

$x=\frac{1}{2-\sqrt{3}}$

$\Rightarrow x=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

$\Rightarrow x=\frac{2+\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}$

$\Rightarrow x=\frac{2+\sqrt{3}}{4-3}$

$\Rightarrow x=2+\sqrt{3} \quad \ldots(1)$

Now,

$x^{2}=(2+\sqrt{3})^{2}$

$\Rightarrow x^{2}=(2)^{2}+(\sqrt{3})^{2}+2(2)(\sqrt{3})$

$\Rightarrow x^{2}=4+3+4 \sqrt{3}$

$\Rightarrow x^{2}=7+4 \sqrt{3} \quad \ldots(2)$

Also,

$x^{3}=x^{2} . x$

$\Rightarrow x^{3}=(7+4 \sqrt{3})(2+\sqrt{3})$

$\Rightarrow x^{3}=14+7 \sqrt{3}+8 \sqrt{3}+12$

$\Rightarrow x^{3}=26+15 \sqrt{3} \quad \ldots(3)$

Now,

$x^{3}-2 x^{2}-7 x+5$

$=(26+15 \sqrt{3})-2(7+4 \sqrt{3})-7(2+\sqrt{3})+5 \quad($ using $(1),(2)$ and $(3))$

$=26+15 \sqrt{3}-14-8 \sqrt{3}-14-7 \sqrt{3}+5$

$=31-28+15 \sqrt{3}-15 \sqrt{3}$

$=3$

Hence, the value of $x^{3}-2 x^{2}-7 x+5$ is 3 .