But they are nested

Let

$a= \sqrt{ \sqrt2 + 2\sqrt{\sqrt2 -1}}$

$b= \sqrt {\sqrt2 - 2\sqrt{\sqrt2-1}}$

We can see that $a^2+b^2= 2 \sqrt2 \qquad...... (1)$

$\begin{aligned} ab &= \sqrt{ \sqrt2 + 2\sqrt{\sqrt2 -1}} \times \sqrt {\sqrt2 - 2\sqrt{\sqrt2-1}} \\ &= \sqrt{2-4 \left ( \sqrt2 -1 \right ) } \\ &= \sqrt{ 6- 4 \sqrt2 } \\ &= \sqrt{ 2^2-2 \times 2 \times \sqrt2 + \left ( \sqrt2 \right )^2 } \\ &= \sqrt{ \left ( 2- \sqrt2 \right )^2 } \\ ab &=2- \sqrt2 \qquad...... (2) \end{aligned}$

Substitute $(1)$ and $(2)$ into the identity $(a+b)^2=a^2+2ab+b^2$ , we get $(a+b)^2=4$

Since $a+b$ is positive, $\large { \boxed{ a+b=2}}$

Letting the given expression be $x,$ we have that

$x^{2} = (\sqrt{2} + 2\sqrt{\sqrt{2} - 1}) + (\sqrt{2} - 2\sqrt{\sqrt{2} - 1}) + 2\sqrt{2 - 4\sqrt{\sqrt{2} - 1}} =$

$2\sqrt{2} + 2\sqrt{6 - 4\sqrt{2}} = 2\sqrt{2} + 2\sqrt{(2 - \sqrt{2})^{2}} = 2\sqrt{2} + 2(2 - \sqrt{2}) = 4.$

Since $x$ is clearly positive, we then conclude that $x = \sqrt{4} = \boxed{2}.$

$\quad\sqrt { \sqrt { 2 } +2\sqrt { \sqrt { 2 } -1 } } +\sqrt { \sqrt { 2 } -2\sqrt { \sqrt { 2 } -1 } } \\ =\sqrt { { \left( \sqrt { \sqrt { 2 } +2\sqrt { \sqrt { 2 } -1 } } +\sqrt { \sqrt { 2 } -2\sqrt { \sqrt { 2 } -1 } } \right) }^{ 2 } } \\ =\sqrt { \sqrt { 2 } +2\sqrt { \sqrt { 2 } -1 } +\sqrt { 2 } -2\sqrt { \sqrt { 2 } -1 } +2\left( \sqrt { \sqrt { 2 } +2\sqrt { \sqrt { 2 } -1 } } \right) \left( \sqrt { \sqrt { 2 } -2\sqrt { \sqrt { 2 } -1 } } \right) } \\ =\sqrt { 2\sqrt { 2 } +2\sqrt { 2-4\left( \sqrt { 2 } -1 \right) } } \\ =\sqrt { 2\sqrt { 2 } +2\sqrt { 6-4\sqrt { 2 } } } \\ =\sqrt { 2\sqrt { 2 } +2\left( 2-\sqrt { 2 } \right) } \\ =\sqrt { 2\sqrt { 2 } +4-2\sqrt { 2 } } \\ =\sqrt { 4 } \\ =\boxed { 2 }$

For answer to be 2 there should be a '-' b/w the two radicals...otherwise the answer would be 2*sqrt(sqrt(2)-1).

simplifying will result to 2