<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] 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.
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