<s>[INST] Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:

$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$

(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
 [/INST] Let the numerator and the denominator 
$$N= \sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\ldots+\sqrt{10+\sqrt{99}}$$
$$D =\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\ldots+\sqrt{10-\sqrt{99}}$$
Apply the denesting formulas 
$$\sqrt{a\pm\sqrt c} = \frac{1}{\sqrt2} \left( \sqrt{a+\sqrt{a^2-c}} \pm 
\sqrt{a-\sqrt{a^2-c}} \right)$$
to get
$$\sqrt{10\pm\sqrt n} = \frac{1}{\sqrt2} \left( \sqrt{10+\sqrt{100-n}} \pm 
\sqrt{10-\sqrt{100-n}} \right)$$
where $n=1,2,...99$. As a result,
$$N= \frac{1}{\sqrt2}(N+D), \>\>\>\>\>D= \frac{1}{\sqrt2}(N-D)$$
Take the ratio,
$$\frac ND=\frac{N+D}{N-D}$$
or,
$$\left(\frac ND\right)^2 - 2\frac ND -1 =0$$
Solve to obtain,
$$\frac ND = 1+\sqrt2$$
</s>