解:$(2) \frac {2}{99\sqrt {97}+97\sqrt {99}}= \frac {2×(99\sqrt {97}-97\sqrt {99})}{99²×97-97²×99}$
$=\frac {2×(99\sqrt {97}-97\sqrt {99})}{99×97×(99-97)}$
$ = \frac {99\sqrt {97}-97\sqrt {99}}{99×97}$
$=\frac {\sqrt {97}}{97}- \frac {\sqrt {99}}{99} $
同理,得$ \frac {2}{3+\sqrt {3}} = 1 - \frac {\sqrt 3}{3} $,$ \frac {2}{5\sqrt {3}+3\sqrt {5}} =\frac {\sqrt {3}}{3}- \frac {\sqrt {5}}{5}$,
$ \frac {2}{7\sqrt {5}+5\sqrt {7}} \frac {\sqrt {5}}{5}-\frac {\sqrt {7}}{7}······$
∴原式$=(1-\frac {\sqrt {3}}{3})+(\frac {\sqrt {3}}{3}-\frac {\sqrt {5}}{5})+(\frac {\sqrt {5}}{5}-\frac {\sqrt {7}}{7})+···+(\frac {\sqrt {97}}{97}-\frac {\sqrt {99}}{99})$
$=1-\frac {\sqrt {99}}{99}=1-\frac {3\sqrt {11}}{99}=\frac {33-\sqrt {11}}{33}$
$(3)$∵$(x + \sqrt {x²+2024}) (y + \sqrt {y²+2024})=2024$
∴$x+ \sqrt {x²+2024}=\frac {2024}{y+\sqrt {y^2+2024}}$
∴$x + \sqrt {x²+2024} =\sqrt {y^2+2024}-y①$
同理,得$y+ \sqrt {y^2+2024}=\sqrt {x²+2024}-x②$
由①+②,得$ x+y+\sqrt {x²+2024}+ \sqrt {y²+2024}= \sqrt {y^2+2024}+\sqrt {x²+2024}-(x+y)$
∴$x+y=0$,即$x+y+2024=2024$