Избавьтесь от иррациональности в знаменателе:

Решите задачу:

$\frac{12}{3+ \sqrt{2}- \sqrt{3}}= \frac{12(3-( \sqrt{2}- \sqrt{3}))}{(3+( \sqrt{2}- \sqrt{3}))( 3 - (\sqrt{2}- \sqrt{3}))}=\frac{12(3-\sqrt{2}+ \sqrt{3})}{3^2 - (\sqrt{2}- \sqrt{3})^2}=\frac{12(3-\sqrt{2}+ \sqrt{3})}{9 - (2-2\sqrt{6}+3)}=$ $\frac{12(3-\sqrt{2}+ \sqrt{3})}{9 - 5+2\sqrt{6}}=\frac{12(3-\sqrt{2}+ \sqrt{3})}{4 +2\sqrt{6}}=\frac{6(3-\sqrt{2}+ \sqrt{3})}{2 +\sqrt{6}}=\frac{6(3-\sqrt{2}+ \sqrt{3})(2- \sqrt{6})}{(2 +\sqrt{6})(2- \sqrt{6})}=$ $\frac{6(6-3 \sqrt{6} -2 \sqrt{2}+\sqrt{12}+2 \sqrt{3}-\sqrt{18})}{4- 6}=\frac{6(6-3 \sqrt{6} -2 \sqrt{2}+2\sqrt{3}+2 \sqrt{3}-3\sqrt{2})}{-2}=$ $-3(6-3 \sqrt{6} -5\sqrt{2}+4\sqrt{3})=9 \sqrt{6} +15 \sqrt{2} -18-12 \sqrt{3}$