解：$(1)$原式$=\frac {3×(\sqrt {10}-\sqrt 7)}{(\sqrt {10}+\sqrt 7)(\sqrt {10}-\sqrt 7)}=\sqrt {10}-\sqrt 7$
原式$=\frac {10-7}{\sqrt {10}+\sqrt 7}=\frac {(\sqrt {10})^2-(\sqrt 7)^2}{\sqrt {10}+\sqrt 7}=\frac {(\sqrt {10}+\sqrt 7)(\sqrt {10}-\sqrt 7)}{\sqrt {10}+\sqrt 7}=\sqrt {10}-\sqrt 7$
$(2) $原式$ =\frac {\sqrt 3-1}{(\sqrt 3+1)(\sqrt 3-1)}+\frac {\sqrt 5-\sqrt 3}{(\sqrt 5+\sqrt 3)(\sqrt 5-\sqrt 3)}+·s+\frac {\sqrt {2 \mathrm n+1}-\sqrt {2 \mathrm n-1}}{(\sqrt {2 \mathrm n+1}+\sqrt {2 \mathrm n-1})(\sqrt {2 \mathrm n+1}-\sqrt {2 \mathrm n-1})}$
$ =\frac {\sqrt 3-1}2+\frac {\sqrt 5-\sqrt 3}2+·s+\frac {\sqrt {2 \mathrm n+1}-\sqrt {2 \mathrm n-1}}2$
$ =\frac {\sqrt {2 \mathrm n+1}-1}2 $
