Hello everybody!

I hope you feel for some algebra!

Problem 1(I have "solved" this one, "solution" below)

$\displaystyle \frac {4}{\sqrt {5} - 3} + 3 + \ \sqrt5 $

Spoiler! - UPDATED

This is how I think

First we put $\displaystyle 3 + \sqrt5$ as a fraction

$\displaystyle \frac {4}{\sqrt {5} - 3} + \frac {3 + \sqrt5} {1} $

Then we multiply $\displaystyle \frac {3 + \sqrt5} {1}$ with $\displaystyle \sqrt {5} - 3$ to get a common denominator.

$\displaystyle \frac {4}{\sqrt {5} - 3} + \frac {(\sqrt {5} - 3)3 + \sqrt5} {\sqrt {5} - 3}$

Ok, now we are close to a more simpler way to put this problem. (Put everything in the same fraction.

$\displaystyle \frac {4 + (\sqrt {5} - 3)3 + \sqrt5} {\sqrt {5} - 3}$

$\displaystyle 4 + (\sqrt {5} - 3)(\sqrt {5} + 3)$ is zero. we are left with

$\displaystyle \frac {0} {\sqrt{5} - 3} = 0 $

End Spoiler

A second opinion on this problem and it's "solution" is always welcome.

Alright,problem 2. Now we got to have a clear mind and a sharp eye to solve this one...

$\displaystyle \frac {a}{ \sqrt{ab} + a} + \frac {b} {\sqrt{ab} - b} - \frac {a}{a-b}$

Solved!

First we work some on the first fraction, to make it more easy to work with. (I'll go more into detail later)

Problem 3$\displaystyle \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}$

Solved!