Question

（2012•黃岡二模）化簡(jiǎn)(

x-y

x2-2xy+y2

-

xy+y2

x2-y2

)•

xy

y-1

=

-

xy

x-y

．

Answer 1

分析：首先對(duì)括號(hào)內(nèi)的每個(gè)分式進(jìn)行化簡(jiǎn)，然后進(jìn)行分式的加減，最后進(jìn)行乘法運(yùn)算即可．

解答：解：原式=[

x-y

(x-y)2

-

y(x+y)

(x+y)(x-y)

]•

xy

y-1

=（

1

x-y

-

y

x-y

）•

xy

y-1

=

1-y

x-y

•

xy

y-1

=-

xy

x-y

．
故答案是：-

xy

x-y

．

點(diǎn)評(píng)：本題主要考查分式的混合運(yùn)算，通分、因式分解和約分是解答的關(guān)鍵．