Избавьтесь от иррациональности в знаменателе:1:1+корень из 2+ корень из 3

$\frac{1}{1+\sqrt{2}+\sqrt{3}} = \frac{1-\sqrt{2}-\sqrt{3}}{(1+\sqrt{2}+\sqrt{3})(1-\sqrt{2}-\sqrt{3})} = \frac{1-\sqrt{2}-\sqrt{3}}{(1+(\sqrt{2}+\sqrt{3}))(1-(\sqrt{2}+\sqrt{3}))} =\\\\ \frac{1-\sqrt{2}-\sqrt{3}}{1-(\sqrt{2}+\sqrt{3})^2} = \frac{1-\sqrt{2}-\sqrt{3}}{1-(2+2\sqrt{2}\sqrt{3}+3)} = \frac{1-\sqrt{2}-\sqrt{3}}{-4-2\sqrt{2}\sqrt{3}} =$

$\frac{(1-\sqrt{2}-\sqrt{3})(4-2\sqrt{2}\sqrt{3})}{-(4+2\sqrt{2}\sqrt{3})(4-2\sqrt{2}\sqrt{3})} = \frac{(1-\sqrt{2}-\sqrt{3})(4-2\sqrt{2}\sqrt{3})}{-(16-4*2*3)} =\\\\ \frac{(1-\sqrt{2}-\sqrt{3})(4-2\sqrt{2}\sqrt{3})}{8} = \frac{4 - 2\sqrt{6} - 4\sqrt{2} + 4\sqrt{3} - 4\sqrt{3} + 6\sqrt{2}}{8} = \frac{4 - 2\sqrt{6} + 2\sqrt{2}}{8}$