若实数x、y满足x2+y2-4x-2y+5=0,则
|
x2+y2-4x-2y+5=0化为x2-4x+4+y2-2y+1=0,
即:(x-2)2+(y-1)2=0,
∴x=2,y=1
∴
=
+yx 3y-2 x
=
+12 3-2 2
=
+12 (
-1)22
=(
+12
-12
+1)2=3+22 2
故答案为:3+2
.2
若实数x、y满足x2+y2-4x-2y+5=0,则
|
x2+y2-4x-2y+5=0化为x2-4x+4+y2-2y+1=0,
即:(x-2)2+(y-1)2=0,
∴x=2,y=1
∴
=
+yx 3y-2 x
=
+12 3-2 2
=
+12 (
-1)22
=(
+12
-12
+1)2=3+22 2
故答案为:3+2
.2