(1)x 2-4x+2的三种配方分别为:
x 2-4x+1=(x-2) 2-3,
x 2-4x+1=(x-1) 2-2x,
(2)由x 2+y 2-4x+6y+13=0得:x 2-4x+4+y 2+6y+9=0,
∴(x-2) 2+(y+3) 2=0
解得:x=2,y=-3
∴2x-y=4+3=7;
(3)a 2+b 2+c 2-ab-3b-2c+4
=(a 2-ab+
1
4 b 2)+(
3
4 b 2-3b+3)+(c 2-2c+1)
=(a 2-ab+
1
4 b 2)+
3
4 (b 2-4b+4)+(c 2-2c+1)
=(a-
1
2 b) 2+
3
4 (b-2) 2+(c-1) 2=0,
从而有a-
1
2 b=0,b-2=0,c-1=0,
即a=1,b=2,c=1,
故a+b+c=4.