问题
解答题
已知x=(
|
答案
由题意知:x=
=1
-12
+1,y=2
=1
+12
-1,2
故x+y=2
,xy=1;x-y=2>0,2y-x=2
-3<0;2
∴原式=x+y x-y
+(x-y)2 y x-2y x(2y-x)2 y
=(x-y)×
+x+y x-y
×y x-2y
×(x-2y)xy y
=x+y+
=2xy
+1.2
已知x=(
|
由题意知:x=
=1
-12
+1,y=2
=1
+12
-1,2
故x+y=2
,xy=1;x-y=2>0,2y-x=2
-3<0;2
∴原式=x+y x-y
+(x-y)2 y x-2y x(2y-x)2 y
=(x-y)×
+x+y x-y
×y x-2y
×(x-2y)xy y
=x+y+
=2xy
+1.2