Question

計(jì)算：

x

x+y

+

y

x-y

-

2xy

x2-y2

．

Answer 1

分析：首先對(duì)分式進(jìn)行通分，然后進(jìn)行同分母的分式的加減運(yùn)算，最后對(duì)結(jié)果進(jìn)行化簡(jiǎn)即可求解．

解答：解：原式=

x(x-y)

x2-y2

+

y(x+y)

x2-y2

-

2xy

x2-y2

=

x(x-y)+y(x+y)-2xy

x2-y2

=

x2+y2-2xy

x2-y2

=

(x-y)2

(x+y)(x-y)

=

x-y

x+y

．

點(diǎn)評(píng)：本題考查了分式的加減，分式的加減運(yùn)算中，如果是同分母分式，那么分母不變，把分子直接相加減即可；如果是異分母分式，則必須先通分，把異分母分式化為同分母分式，然后再相加減．