Simplifying $\left(\frac{a-b}{\sqrt a+\sqrt b}+\sqrt b\right)/\left(\frac{a-b}{\sqrt a+\sqrt b}-\sqrt a\right)$ It was really a long time since I did math at school and now I'm preparing for a admission test to college.
I took a look at the older tests and immediately got stuck at the first question. It really feels like I have to shape up.
Anyway so here's the equation:

$$x = \frac{\frac{a-b}{\sqrt a+\sqrt b}+\sqrt b}{\frac{a-b}{\sqrt a+\sqrt b}-\sqrt a}$$

...and it is supposed to end up in $-\sqrt\frac{a}{b}$
I tried a few different ways of doing it, but always end up wrong. I don't wanna post my attempts here since I honestly don't seem to know what I'm doing. Also this is my first post here on math.stackexchange.com so please tell me if I'm doing something wrong. Any help is greatly appreciated (with the problem).
 A: Notice that $$a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})$$
so you get $$(\sqrt{a}-\sqrt{b}+\sqrt{b})/(\sqrt{a}-\sqrt{b}-\sqrt{a})=-\sqrt{a\over b}$$
A: Hint:
Multiply the denominator and numerator by $\sqrt{a} + \sqrt b$. Then, tell us what you get, and we can talk further.
A: Let us start by simplifying the numerator and denominator separately:
$$num = \frac{a-b}{\sqrt a+\sqrt b}+\sqrt{b} = \frac{a-b}{\sqrt a+\sqrt b} + \frac{\sqrt{b}({\sqrt a+\sqrt b})}{{\sqrt a+\sqrt b}}$$ by dividing and multiplying $\sqrt{b}$ by the denominator of the first term. Now we expand and do the maths:
$$num = \frac{a-b+\sqrt{b}{(\sqrt{a}+\sqrt{b})}}{\sqrt{a}+\sqrt{b}} = \frac{a-b+\sqrt{ab}{+b}}{\sqrt{a}+\sqrt{b}} = \frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$$
By a similar process we get that the denominator is
$$den = \frac{a-b}{\sqrt a+\sqrt b}-\sqrt{a} = \frac{-b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$$
Now when you put them back together, you use the following rule: $$\frac{\frac{a}{d}}{\frac{b}{d}} = \frac{a}{b}$$ to get
$$x = \frac{a+\sqrt{ab}}{-b-\sqrt{ab}}$$
Now try spliting $a$ into $\sqrt{a}\sqrt{a}$ and $-b$ into $-\sqrt{b}\sqrt{b}$ and use some algebraic manipulation to get to the final result.
EDIT: as per request, I will go further. (Notice that I had a very unfortunate sign flipped on the simplification of the denominator above)
$a = \sqrt{a}\sqrt{a}$ and $-b = -\sqrt{b}\sqrt{b}$ so that, substituting back,
$$x = \frac{\sqrt{a}\sqrt{a}+\sqrt{ab}}{-\sqrt{b}\sqrt{b}-\sqrt{ab}} = \frac{\sqrt{a}(\sqrt{a} + \sqrt{b})}{-\sqrt{b}(\sqrt{b} + \sqrt{a})} = \frac{\sqrt{a}}{-\sqrt{b}} = -\sqrt{\frac{a}{b}}$$
A: just simplify the equation:
$ \Large x =\frac{\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)}+\sqrt{b}}{\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)}-\sqrt{a}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)-\sqrt{a}}=\frac{\sqrt{a}}{-\sqrt{b}}=-\sqrt{\frac{a}{b}}$
A: Doing the common denominator and simplifying, I obtain: $\frac{a+\sqrt{ab}}{-(b+\sqrt{ab})}$. From this, I multiply the numerator and the denominator by $b-\sqrt{ab}$ and I get: $\frac{(a+\sqrt{ab})(b-\sqrt{ab})}{-(b+\sqrt{ab})(b-\sqrt{ab})}=\frac{ab-a\sqrt{ab}+b\sqrt{ab}-ab}{-(b^2-b\sqrt{ab}+b\sqrt{ab}-ab)}=\frac{\sqrt{ab}(b-a)}{-b(b-a)}=-\frac{\sqrt{ab}}{b}=-\sqrt{\frac{ab}{b^2}}=-\sqrt{\frac{a}{b}}$
