If $$x=\frac{1}{\sqrt2+1}$$​, then what will be the value of $$x+1$$​?

Correct option is A

Given:

x = $$\frac{1}{\sqrt{2} + 1}$$​
Solution:

​$$x = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$$​​

​$$x + 1 = \sqrt{2} - 1 + 1 = \sqrt{2}$$​