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$\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}=2\sqrt{x^3}\\\ x \neq +-1\\\\ \frac{\sqrt{x}+1}{\sqrt{x}-1} - \frac{\sqrt{x}-1}{\sqrt{x}+1} = 2x\sqrt{x}\\\\\ \sqrt{x}+1=t(\sqrt{x}-1)\\ \\ t\sqrt{x}-\sqrt{x}=t+1\\\\ \sqrt{x}=\frac{t+1}{t-1}\\\\ t-\frac{1}{t}=2*\frac{(t+1)^3}{(t-1)^3}\\\\ (t^2-1)(t-1)^3=2t(t+1)^3\\\\ (t+1)(t-1)^4=2t(t+1)^3\\\\ (t+1)((t-1)^4-2t(t+1)^2)=0\\ t=-1\\ t=3-\sqrt{8}; t=\sqrt{8}-3\\$
$t=-1\\ t=3-\sqrt{8}\\ t=\sqrt{8}+3\\\\ \frac{\sqrt{x}+1}{\sqrt{x}-1} = -1\\ \sqrt{x}+1=1-\sqrt{x}\\ x=0\\\\ \sqrt{x}+1 = (\sqrt{8}+3)\sqrt{x}-\sqrt{8}-3\\\\ x=2$
Ответ $x=0;2$