Question

當(dāng)x=

1

2 -1

，y=

1

2 +1

時，求下述代數(shù)式的值．
(x+y)

x2

x2-y2

+

y2

x-y

-(2x2y-2xy2)÷(x2-2xy+y2)．

Answer 1

分析：本題的關(guān)鍵是化簡，然后把給定的值代入計算．

解答：解：原式=

x2

x-y

+

y2

x-y

-

2xy(x-y)

(x-y)2

=

x2+y2

x-y

-

2xy

x-y

=

(x-y)2

x-y

=x-y，
當(dāng)x=

1

2 -1

，y=

1

2 +1

時，
原式=

1

2 -1

-

1

2 +1

=

2

+1-（

2

-1）=2．

點(diǎn)評：本題考查了分式的計算和化簡．解決這類題目關(guān)鍵是把握好通分與約分．分式加減的本質(zhì)是通分，乘除的本質(zhì)是約分．同時注意在進(jìn)行運(yùn)算前要盡量保證每個分式最簡．