问题 填空题
用配方法使下面等式成立:
(1)x2-2x-3=(x-______)2-______;
(2)x2+0.4x+0.5=(x+______)2+______;
(3)3x2+2x-2=3(x+______)2+______;
(4)
2
3
x2+
1
3
x-2=
2
3
(x+______)2+______.
答案

(1)x2-2x-3=x2-2x+1-4=(x-1)2-4;

故答案为:1,4

(2)x2+0.4x+0.5=x2+0.4x+0.04+0.46=(x+0.2)2+0.46;

故答案为:0.2,0.46

(3)3x2+2x-2=3(x2+

2
3
x)-2=3(x2+
2
3
x+
1
9
)-
7
3
=3(x+
1
3
2-
7
3

故答案为:

1
3
-
7
3

(4)

2
3
x2+
1
3
x-2=
2
3
(x2+
1
2
x)-2=
2
3
(x2+
1
2
x+
1
4
)-
49
24
=
2
3
(x+
1
4
2-
49
24

故答案为:

1
4
-
49
24

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