<s>[INST] 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: 
$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}}=\sqrt{x}\Leftrightarrow \frac{\left( \sqrt{x}-\sqrt{x+\sqrt{3}} \right)\left( x+\sqrt{3} \right)}{-\sqrt{3}}+\frac{\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right)\left( x-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}$$
$$\frac{2\sqrt{x}\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)}{\sqrt{3}}=\sqrt{x}\Leftrightarrow 2\left( \sqrt{x}\sqrt{x+\sqrt{3}}-\sqrt{3} \right)=\sqrt{3}\Leftrightarrow \sqrt{x}\sqrt{x+\sqrt{3}}=3\sqrt{3}$$
 [/INST] 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 $\sqrt{x + \sqrt{3}} \neq \sqrt{x - \sqrt{3}}$.)
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