试题

题目:
(2002·娄底)当x=
1
2
-1
,y=
1
2
+1
时,求下述代数式的值.
(x+y)
x2
x2-y2
+
y2
x-y
-(2x2y-2xy2)÷(x2-2xy+y2)

答案
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,
当x=
1
2
-1
,y=
1
2
+1
时,
原式=
1
2
-1
-
1
2
+1
=
2
+1-(
2
-1)=2.
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,
当x=
1
2
-1
,y=
1
2
+1
时,
原式=
1
2
-1
-
1
2
+1
=
2
+1-(
2
-1)=2.
考点梳理
分式的化简求值;分母有理化.
本题的关键是化简,然后把给定的值代入计算.
本题考查了分式的计算和化简.解决这类题目关键是把握好通分与约分.分式加减的本质是通分,乘除的本质是约分.同时注意在进行运算前要尽量保证每个分式最简.
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