Determine the value of 'x' in $$x=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+2}$$

Expression : $$x=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+2}$$

Rationalizing the denominator, we get :

= $$(\frac{1}{1+\sqrt{2}}\times\frac{\sqrt2-1}{\sqrt2-1})+(\frac{1}{\sqrt{2}+\sqrt{3}}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2})+(\frac{1}{\sqrt{3}+2}\times\frac{2-\sqrt3}{2-\sqrt3})$$

Using, $$(a+b)(a-b)=a^2-b^2$$

= $$\frac{\sqrt2-1}{(2-1)}+\frac{\sqrt3-\sqrt2}{3-2}+\frac{2-\sqrt3}{4-3}$$

= $$(\sqrt2-1)+(\sqrt3-\sqrt2)+(2-\sqrt3)$$

= $$2-1=1$$

=> Ans - (C)