Solve the irrational radical equation Solve equation:$$\frac{x + \sqrt{3}}{\sqrt{x} + \sqrt{x + \sqrt{3}}} + \frac{x - \sqrt{3}}{\sqrt{x} - \sqrt{x - \sqrt{3}}} = \sqrt{x}$$
answer: $S=\{2\}$
Attemp: 
$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}}=\sqrt{x}\Leftrightarrow \frac{\left( \sqrt{x}-\sqrt{x+\sqrt{3}} \right)\left( x+\sqrt{3} \right)}{-\sqrt{3}}+\frac{\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right)\left( x-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}$$
$$\frac{2\sqrt{x}\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}\Leftrightarrow 2\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)=\sqrt{3}\Leftrightarrow \sqrt{x}\sqrt{x+\sqrt{3}}=3\sqrt{3}$$
 A: You have a mistake in your algebra (see the previous answer) but rationalizing the denominator is a good idea. Note that the domain is $x\geq \sqrt 3$. After you do that and multiply both sides by $\sqrt{3}$, you obtain
$$x\sqrt{x-\sqrt{3}}-\sqrt{3}\sqrt{x-\sqrt{3}}+x\sqrt{x+\sqrt{3}}+\sqrt{3}\sqrt{x+\sqrt{3}}=3\sqrt 3\sqrt x $$ which can be rewritten as 
$$(x-\sqrt3)^{3/2}+(x+\sqrt3)^{3/2}=3\sqrt 3\sqrt x $$
Then, square both sides (since they are positive, squaring produces an equivalent equation):
$$(x-\sqrt3)^{3}+(x+\sqrt3)^{3}+2(x^2-3)^{3/2}=27x \Leftrightarrow \\
2x^3+2(x^2-3)^{3/2}=9x  $$
If you take the derivative of $f(x)=2x^3+2(x^2-3)^{3/2}-9x$ and use that $x\geq \sqrt 3$, it follows that $f'(x)>0$. This means that $f(x)$ is strictly increasing on the interval $x\geq \sqrt 3$, and the equation has no more than $1$ root. We can check $x=2$ is a root, so that's the final answer.
A: When you inserted factors to obtain a common denominator, you made the error
$$\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right) \left( \sqrt{x}-\sqrt{x-\sqrt{3}} \right) \neq -\sqrt{3} \text{.} $$
That you did not get the same denominator in both fractions should have highlighted a problem.
Since we require $x - \sqrt{3} \geq 0$ for the original expression to be defined, we may assume $x \geq \sqrt{3}$, which fact we will use in combining radicals.
$$\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right) \left( \sqrt{x}-\sqrt{x-\sqrt{3}} \right)  \\
= x + \sqrt{x^2+x\sqrt{3}} - \sqrt{x^2-x\sqrt{3}} - \sqrt{x^2 - 3}  \text{.}  $$
(This is not an instance of the difference of two squares factorization because $\sqrt{x + \sqrt{3}} \neq \sqrt{x - \sqrt{3}}$.)
