电子课本网 第42页

第42页

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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$​