Impossible Square Root

I'll be working out on a general expression that is by taking $2005=x$ in the given expression.

So, we try to simplify the given expression by rationalizing as follows : $\begin{aligned} \frac{1}{\sqrt{x+\sqrt{x^2-1}}} & = & \sqrt{\frac{1}{x+\sqrt{x^2-1}}} \\ & = & \sqrt{\frac{1}{x+\sqrt{x^2-1}}\times\frac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}} \\ & = & \sqrt{\frac{x-\sqrt{x^2-1}}{x^2-(x^2-1)}} \\ & = & \sqrt{x-\sqrt{x^2-1}} \\ \end{aligned}$

Now, here we try a little bit of manipulation. $\begin{aligned} \sqrt{x-\sqrt{x^2-1}} & = & \sqrt{\frac{(x+1)+(x-1)}{2}-\sqrt{(x+1)(x-1)}} \\ \end{aligned}$

To simplify the work, I'll be using the following substitutions :

$a = \sqrt{x+1}$ and $b = \sqrt{x-1}$ .

Now, plugging this into our equation, we get $\begin{aligned} \sqrt{\frac{(x+1)+(x-1)}{2}-\sqrt{(x+1)(x-1)}} & = & \sqrt{\frac{a^2+b^2}{2}-ab} \\ & = & \sqrt{\frac{a^2+b^2-2ab}{2}} \\ & = & \sqrt{\frac{(a-b)^2}{2}} \\ & = & \frac{|a-b|}{\sqrt{2}} \\ \end{aligned}$

So, we are done and now after substituting back the values of $a,b$ (also notice that we have $a>b$ ) , we get

$\begin{aligned} \frac{|a-b|}{\sqrt{2}} & = & \sqrt{\frac{x+1}{2}}-\sqrt{\frac{x-1}{2}} \end{aligned}$

And thus, $\frac{1}{\sqrt{x+\sqrt{x^2-1}}}=\sqrt{\frac{x+1}{2}}-\sqrt{\frac{x-1}{2}}$

Substituting $x=2005$ gives our desired answer $\boxed{\frac{1}{\sqrt{2005+\sqrt{2005^2-1}}}=\sqrt{1003}-\sqrt{1002}}$

$\displaystyle \begin{aligned} \frac {1}{\sqrt{2005+\sqrt{2005^2-1}}} & = \frac {\sqrt{2005-\sqrt{2005^2-1}}}{\sqrt{2005+\sqrt{2005^2-1}}\sqrt{2005-\sqrt{2005^2-1}}} \\ & = \sqrt{2005-\sqrt{2005^2-1}} \\ & = \sqrt{2005-\sqrt{(2005-1)(2005+1)}} \\ & = \sqrt{2005-\sqrt{(2004)(2006)}} \\ & = \sqrt{(\sqrt{1003})^2 -2\sqrt{1002} \sqrt{1003} + (\sqrt{1002})^2} \\ & = \sqrt{\left( \sqrt{1003} - \sqrt{1002}\right)^2} \\ & = \boxed {\sqrt{1003} - \sqrt{1002}} \end{aligned}$

Let us solve by rationalizing the denominator
$\displaystyle \dfrac{1}{\sqrt{x+\sqrt{x^2-1}}}* \dfrac{\sqrt{x-\sqrt{x^2-1}}}{\sqrt{x-\sqrt{x^2-1}}}=\dfrac{\sqrt{x-\sqrt{x^2-1}}}{\sqrt1}\\=\sqrt{x-\sqrt{(x+1)(x-1)}} \\=\sqrt{ \{\sqrt{ \dfrac{x+1}{2} }\}^2 +\{\sqrt{ \dfrac{x-1}{2} }\}^2 -2*\{\sqrt{ \dfrac{x+1}{2} }\}*\{\sqrt{ \dfrac{x-1}{2} }\}}\\= \sqrt{ \{ \sqrt{ {\dfrac{x+1}{2} }}-\sqrt{ \dfrac{x-1}{2} } }\}^2\\= \large \sqrt{ {\dfrac{x+1}{2} }}-\sqrt{ \dfrac{x-1}{2} }$
$If~x= 2005,~~\sqrt{\dfrac {2005+ 1 }{2}} -\sqrt { \dfrac{ 2005-1 }{ 2 } } \\= \sqrt{ 1003}-\sqrt{1002 }$

Heads up: my solution is just how I prefer to solve this problem... there are other thorough solutions below. Let's tackle the denominator as is. Since all of the answer choices are a difference between two radical expressions, we know that the denominator will most likely factor to a sum of radical expressions. Let's let the larger of these radical expressions equal \sqrt{a} and the smaller one equal \sqrt{b} . Now, we can set an equation: \sqrt{a} + \sqrt{b} = \sqrt{2005 + \sqrt{2005^2 - 1}} We can continue by squaring both sides: a + b + 2 \sqrt{ab} = 2005 + \sqrt{2005^2 - 1} From this equation, we can infer that a + b = 2005 and 2\sqrt{ab} = \sqrt{2005^2 - 1} (Remember that the radicands of the answer choices are all integers.) Now, we tackle 2\sqrt{ab} = \sqrt{2005^2 - 1}. Squaring both sides yields 4ab = 2005^2 - 1 = (2004)(2006) ( Difference of squares expansion ) We find that ab = (2006)(501). By inspection, we can see that a = 1003 and b = 1002 solve our system of equations. YOU'RE ALMOST THERE! Now, we have \frac{1}{\sqrt{1003} + \sqrt{1002}}. Rationalizing the denominator, we get \boxed{\sqrt{1003} - sqrt{1002}} :D