How to show the expansion of $(a-b)(a+b)$ Now the actual question is to show $\sqrt{x} - \sqrt{2} = \frac{x-2}{\sqrt{x}  + \sqrt{2}}$. I know it is obvious but I am not sure how to show it in a way that is mathematically correct.  
Also, it asks me to deduce that $∣\sqrt{x} - \sqrt{2}∣ \le \frac{∣x-2∣}{\sqrt{2}}$ . Now again, this seems obvious to me because $\sqrt{x}$ is always positive therefore $\sqrt{2} \le \sqrt{x}  + \sqrt{2}$, and when you divide by a smaller number, the total fraction increases. Again, I am not sure how to word it so it is sophisticated. 
 A: To answer your first question, it's a simple matter of multiplying  the expression by the numerator's conjugate over itself.
$$(\sqrt x -\sqrt 2) \;\cdot\frac{\sqrt x +\sqrt 2}{\sqrt x +\sqrt 2} = \frac{x-2}{\sqrt x +\sqrt 2}$$
To answer your second, I am assuming that x is nonnegative to avoid imaginary numbers. We know from the first question that $\sqrt x -\sqrt 2=\frac{x-2}{\sqrt x +\sqrt 2}$. Now let's look at that inequality again.
$$\frac{|x-2|}{\sqrt 2}\ge\frac{|x-2|}{\sqrt x +\sqrt 2}$$
The only difference between the two expressions is the $\sqrt x$ in the denominator. No matter the $x$ value assuming $x \in \Bbb R$, we get zero or a positive number for $\sqrt x$. Thus, our denominator is greater than or equal to $\sqrt 2$, and consequently our rightmost expression is--at most--equal to the leftmost expression.
A: To solve the first one we need to multiply by the conjugate of the expression.  The conjugate of an expression $ax - b$ is $ax + b$ or the reverse.
In your equation we the conjugate is $\sqrt x + \sqrt 2$.  We can now multiply $\sqrt x - \sqrt 2$ by $(\sqrt x + \sqrt 2)/(\sqrt x + \sqrt 2)$ or 1.  When you do this you have to be careful that the denominator is never zero but you are fine in this case because $\sqrt x$ can never be negative.
When we multiple this out we get the answer you are looking for: $(x-2)/((\sqrt x + \sqrt 2)$.
The reason we do this is to make certain expressions easier to manipulate or to simplify unusual expressions.  It works because $(a^2-b^2) = (a-b)*(a+b)$.
As for the inequality your explanation is great.
A: The FOIL (first outside inside last) method:
$$
(a-b)(a+b) = aa+ab-ba-bb
$$
or for your "actual" question, multiply both sides by $\sqrt{x} + \sqrt{2}$ and:
$$
(\sqrt{x} - \sqrt{2})(\sqrt{x} + \sqrt{2}) = x + \sqrt{2x} - \sqrt{2x} - 2 = x-2
$$
