Освободитесь от иррациональности в знаменатели дроби

Решите задачу:

$\frac{4}{ \sqrt{3} +1} = \frac{4( \sqrt{3} -1)}{( \sqrt{3}+1)( \sqrt{3} -1)} = \frac{4( \sqrt{3} -1)}{3-1} = \frac{4( \sqrt{3} -1)}{2} =2( \sqrt{3} -1)$
$\frac{1}{1- \sqrt{2} } = \frac{1+ \sqrt{2} }{(1- \sqrt{2})(1+ \sqrt{2} ) } = \frac{1+ \sqrt{2} }{1-2} =-1- \sqrt{2}$
$\frac{4}{ \sqrt{10}- \sqrt{2}}= \frac{4( \sqrt{10}+ \sqrt{2} )}{( \sqrt{10}+ \sqrt{2} )( \sqrt{10}- \sqrt{2} )} = \frac{4( \sqrt{10}+ \sqrt{2} )}{10-2} = \frac{4( \sqrt{10}+ \sqrt{2} )}{8}= \frac{ \sqrt{10}+ \sqrt{2} }{2}$
$\frac{12}{ \sqrt{3} + \sqrt{6}} = \frac{12(\sqrt{3}-\sqrt{6})}{(\sqrt{3} + \sqrt{6})(\sqrt{3} - \sqrt{6})} =\frac{12(\sqrt{3}-\sqrt{6})}{3-6} =\frac{12(\sqrt{3}-\sqrt{6})}{-3} =-4( \sqrt{3}- \sqrt{6})= \\ =4( \sqrt{6} - \sqrt{3} )$