How to prove the equality $|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$ How to prove this equality?
$|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$
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 A: Hint:
Note that in your equality $\sqrt{x}+\sqrt{a}>0$ and if $n>0$, then $n|m|=|nm|$. So, multiply both sides by the positive quantity $\sqrt{x}+\sqrt{a}$ and simplify using the formula for the difference of squares: $x^2-y^2=(x-y)(x+y)$.
Alternatively, you can arrive at the right-hand side of the equality by multiplying $|\sqrt{x}-\sqrt{a}|$ by $\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}$:
$$
|\sqrt{x}-\sqrt{a}|=
|\sqrt{x}-\sqrt{a}|\cdot 1=
|\sqrt{x}-\sqrt{a}|\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}=\\
\frac{|(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})|}{\sqrt{x}+\sqrt{a}}=
\frac{|x-a|}{\sqrt{x}+\sqrt{a}}.
$$
Of course, all this is true as long as $\sqrt{x}+\sqrt{a}\ne0$.
A: Assuming both $a$ and $x$ are positive.  Two cases (1) $x \gt a$ and (2) $x\le a$.
Case(1) becomes $\sqrt{x}-\sqrt{a}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$.  Clear the denominator and get $x-a=x-a$.  Since $x\gt a$, $x-a=|x-a|$
Case(2) becomes $\sqrt{a}-\sqrt{x}=\frac{a-x}{\sqrt{x}+\sqrt{a}}$.  Clear the denominator and get $a-x-a-x$. Since $x\le a$, $a-x=|a-x|=|x-a|$.
