If $$x=\sqrt{1+\frac{\sqrt{3}}{2}-}\sqrt{1-\frac{\sqrt{3}}{2}}$$, then the value of $$\frac{\sqrt{2}-x}{\sqrt{2}+x}$$ will be closest to:

$$x=\sqrt{1+\frac{\sqrt{3}}{2}-}\sqrt{1-\frac{\sqrt{3}}{2}}$$
$$x^2 =(\sqrt{1+\frac{\sqrt{3}}{2}-}\sqrt{1-\frac{\sqrt{3}}{2}})^2$$
$$x^2 =1+\frac{\sqrt{3}}{2} + 1-\frac{\sqrt{3}}{2} - 2\sqrt{1+\frac{\sqrt{3}}{2}} \sqrt{1-\frac{\sqrt{3}}{2}}$$
$$x^2 =1+\frac{\sqrt{3}}{2} + 1-\frac{\sqrt{3}}{2} - 2\sqrt{1+\frac{\sqrt{3}}{2}} \sqrt{1-\frac{\sqrt{3}}{2}}$$
$$x^2 =1+\frac{\sqrt{3}}{2} + 1-\frac{\sqrt{3}}{2} - 2\sqrt{1- \frac{3}{4}} $$
$$x^2 = 2 - 2\sqrt{ \frac{1}{4}} $$
$$x^2 = 1$$
x = 1 or x = -1
Now,
$$\frac{\sqrt{2}-x}{\sqrt{2}+x}$$
= $$\frac{\sqrt{2}-x}{\sqrt{2}+x} \times \frac{\sqrt{2}-x}{\sqrt{2}-x}$$
= $$\frac{2 - 2\sqrt{2}x + x^2}{2 - x^2}$$
Put the value of x = 1,
= $$\frac{2 - 2\sqrt{2} + 1}{2 - 1}$$
= $$3 - 2\sqrt{2}$$ = 0.17