Решите неравенство: (10-2корень из 21)x>корень из 7-корень из 3.

\[\left( 10 - 2\sqrt{21} \right)x > \sqrt{7} - \sqrt{3}\]

\[\left( \sqrt{7} - \sqrt{3} \right)^{2}x > \sqrt{7} - \sqrt{3}\]

\[\left( \sqrt{7} - \sqrt{3} \right)^{2}x - \left( \sqrt{7} - \sqrt{3} \right) > 0\]

\[\left( \sqrt{7} - \sqrt{3} \right)x - 1 > 0\]

\[\left( \sqrt{7} - \sqrt{3} \right)x > 1\]

\[x > \frac{1}{\sqrt{7} - \sqrt{3}}\]

\[\frac{1}{\sqrt{7} - \sqrt{3}} =\]

\[= \frac{\sqrt{7} + \sqrt{3}}{\left( \sqrt{7} - \sqrt{3} \right)\left( \sqrt{7} + \sqrt{3} \right)} =\]

\[= \frac{\sqrt{7} + \sqrt{3}}{7 - 3} = \frac{\sqrt{7} + \sqrt{3}}{4}\]

\[x > \frac{\sqrt{7} + \sqrt{3}}{4}\]

\[Ответ:\left( \frac{\sqrt{7} + \sqrt{3}}{4}; + \infty \right).\]