Solving the equation $\frac{x+\sqrt 3}{\sqrt x + \sqrt {x+\sqrt 3}} + \frac{x-\sqrt 3}{\sqrt x - \sqrt {x-\sqrt 3}} = \sqrt x$ for $x$ I have the following equation: 
$$\frac{x+\sqrt 3}{\sqrt x + \sqrt {x+\sqrt 3}} + \frac{x-\sqrt 3}{\sqrt x - \sqrt {x-\sqrt 3}} = \sqrt x$$
I know that $x=2$, but I don't know steps to get there.

Take lcm & simplify as follows $$\frac{x+\sqrt  3}{\sqrt x+\sqrt{x+\sqrt 3}}+\frac{x-\sqrt  3}{\sqrt x-\sqrt{x-\sqrt 3}}=\sqrt x$$
$$\frac{(x+\sqrt  3)(\sqrt x-\sqrt{x-\sqrt 3})+(x-\sqrt 3)(\sqrt x+\sqrt{x+\sqrt 3})}{(\sqrt x+\sqrt{x+\sqrt 3})(\sqrt x-\sqrt{x-\sqrt 3})}=\sqrt x$$
$$2x\sqrt x-x\sqrt{x-\sqrt 3}-\sqrt 3\sqrt{x-\sqrt 3}+x\sqrt{x+\sqrt 3}-\sqrt 3\sqrt{x+\sqrt 3}$$$$=x\sqrt x+x\sqrt{x+\sqrt 3}-x\sqrt x-x\sqrt{x-\sqrt 3}-\sqrt x\sqrt{x^2-3}$$
$$x\sqrt x+\sqrt x\sqrt{x^2-3}-\sqrt 3\left(\sqrt{x+\sqrt 3}+\sqrt{x-\sqrt 3}\right)=0$$
re-arranging & taking squares on both the sides,
$$\left(\sqrt x(x+\sqrt{x^2-3})\right)^2=\left(\sqrt 3\left(\sqrt{x+\sqrt 3}+\sqrt{x-\sqrt 3}\right)\right)^2$$
$$2x^3-9x=(2x^2-6)\sqrt{x^2-3}$$
$$2x^3-9x=2(x^2-3)^{3/2}$$
taking squares on both the sides & simplifying, one should get 
$$27x^2-108=0$$
$$x^2=4\iff x=\pm 2$$
but $\sqrt x$ is undefined for $x<0$ hence, the correct value is $$\boxed{\color{red}{x=2}}$$
