x=

\sqrt{10+\sqrt 1} = \frac{1}{\sqrt 2} \left(\sqrt{10+\sqrt{99}} + \sqrt{10-\sqrt{99}}\right)

and

\sqrt{10-\sqrt 1} = \frac{1}{\sqrt 2} \left(\sqrt{10+\sqrt{99}} - \sqrt{10-\sqrt{99}}\right)

Thus the given expression can be written as

\frac{a+b}{a-b}

where

a = \sqrt{10+\sqrt 1} +\sqrt{10+\sqrt 2}+...+\sqrt{10+\sqrt {99}} and

b = \sqrt{10-\sqrt 1} +\sqrt{10-\sqrt 2}+...+\sqrt{10-\sqrt {99}}

so that we have

\frac{a}{b} = \frac{a+b}{a-b} \Rightarrow a^2 - 2ab -b^2 =0 \Rightarrow \frac{a}{b} = \boxed{\sqrt 2+1}

Simply Amazing Solution !

Yes Aditiya Great Solution

Thanks hsbhatt Sir