Sure.. Square both sides.. v2.0

let $x + \sqrt{x} = a$ and $x - \sqrt{x} = b$

thus $a - b = 2\sqrt{x}$

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = 2\sqrt{x}$

$\sqrt{a} + \sqrt{b} = \dfrac{200\sqrt{x + \sqrt{x}}}{199}$

$\sqrt{a} - \sqrt{b} = \dfrac{199\sqrt{x}}{100\sqrt{x + \sqrt{x}}}$

thus adding both the equations

we get

$2\sqrt{x + \sqrt{x}} = \dfrac{200\sqrt{x + \sqrt{x}}}{199} + \dfrac{199\sqrt{x}}{100\sqrt{x + \sqrt{x}}}$

thus

$39800( x + \sqrt{x}) = 20000(x + \sqrt{x}) + 39601\sqrt{x}$

$19800(x + \sqrt{x}) = 39601\sqrt{x}$

$19800x = 19801\sqrt{x}$

$\sqrt{x}(19800\sqrt{x} - 19801) = 0$

as x can't be zero thus

$\sqrt{x} = \dfrac{19801}{19800}$

we can easily see that denominator is just 1 lesser than numerator thus they are co - prime, thus their squares will also be co-prime.

$x = \dfrac{392079601}{392040000}$

$m + n = 784119601$

$\sqrt{x+\sqrt{x}} - \sqrt{x-\sqrt{x}} = \dfrac {199}{100} \sqrt {\dfrac {x} {x+\sqrt{x}} }$

Multiply throughout by $\sqrt{x+\sqrt{x}}:\quad \Rightarrow x + \sqrt {x} - \sqrt {x^2 - x } = 1.99 \sqrt {x}$

$\Rightarrow x - 0.99 \sqrt {x} = \sqrt {x^2 - x }$

Squaring both sides: $\quad \Rightarrow x^2 - 1.98x\sqrt {x} + 0.9801x = x^2 - x$

$\Rightarrow - 1.98x\sqrt {x} + 0.9801x = - x$

Dividing by $x$ throughout and rearrange: $\quad \Rightarrow \sqrt {x} = \dfrac {1.9801}{1.98}$

$\Rightarrow x = \left( \dfrac {19801}{19800} \right) ^2 = \dfrac {392079601} {392040000} = \dfrac {m}{n} \quad \Rightarrow m+n = \boxed {784119601}$

{Reduce[-Sqrt[-Sqrt[x] + x] + Sqrt[Sqrt[x] + x] == (199 Sqrt[x/(Sqrt[x] + x)])/100, x], N[Reduce[-Sqrt[-Sqrt[x] + x] + Sqrt[Sqrt[x] + x] == (199 Sqrt[x/(Sqrt[x] + x)])/100, x]]}