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用配方法解方程13x2−x−4=0,配方后得()A.(x−32)2=394B.(x−32)2=−394C.(x−32)2=574D.(x−32)2=12
题目详情
用配方法解方程
x2−x−4=0,配方后得( )
A.(x−
)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12
x2−x−4=0,配方后得( )
A.(x−
)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12
1 1 3 3 x2−x−4=0,配方后得( )
A.(x−
)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12x2−x−4=0,配方后得( )
A.(x−
)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=122−x−4=0,配方后得( )
A.(x−
)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12
(x−
)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12
3 3 2 2 )2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12)2=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=122=
B.(x−
)2=−
C.(x−
)2=
D.(x−
)2=12
39 39 4 4
(x−
)2=−
C.(x−
)2=
D.(x−
)2=12
3 3 2 2 )2=−
C.(x−
)2=
D.(x−
)2=12)2=−
C.(x−
)2=
D.(x−
)2=122=−
C.(x−
)2=
D.(x−
)2=12
39 39 4 4
(x−
)2=
D.(x−
)2=12
3 3 2 2 )2=
D.(x−
)2=12)2=
D.(x−
)2=122=
D.(x−
)2=12
57 57 4 4
(x−
)2=12
3 3 2 2 )2=12)2=122=12
1 |
3 |
A.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
1 |
3 |
A.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
1 |
3 |
A.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
A.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
A.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
39 |
4 |
B.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
39 |
4 |
(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
39 |
4 |
C.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
39 |
4 |
(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
57 |
4 |
D.(x−
3 |
2 |
57 |
4 |
(x−
3 |
2 |
3 |
2 |
▼优质解答
答案和解析
∵
x2−x−4=0
∴x2-3x-12=0
∴x2-3x=12
∴x2-3x+
=12+
∴(x-
)2=
故选C.
1 1 13 3 3x2−x−4=0
∴x2-3x-12=0
∴x2-3x=12
∴x2-3x+
=12+
∴(x-
)2=
故选C. 2−x−4=0
∴x22-3x-12=0
∴x22-3x=12
∴x22-3x+
=12+
∴(x-
)2=
故选C.
9 9 94 4 4=12+
∴(x-
)2=
故选C.
9 9 94 4 4
∴(x-
)2=
故选C.
3 3 32 2 2)22=
故选C.
57 57 574 4 4
故选C.
1 |
3 |
∴x2-3x-12=0
∴x2-3x=12
∴x2-3x+
9 |
4 |
9 |
4 |
∴(x-
3 |
2 |
57 |
4 |
故选C.
1 |
3 |
∴x2-3x-12=0
∴x2-3x=12
∴x2-3x+
9 |
4 |
9 |
4 |
∴(x-
3 |
2 |
57 |
4 |
故选C. 2−x−4=0
∴x22-3x-12=0
∴x22-3x=12
∴x22-3x+
9 |
4 |
9 |
4 |
∴(x-
3 |
2 |
57 |
4 |
故选C.
9 |
4 |
9 |
4 |
∴(x-
3 |
2 |
57 |
4 |
故选C.
9 |
4 |
∴(x-
3 |
2 |
57 |
4 |
故选C.
3 |
2 |
57 |
4 |
故选C.
57 |
4 |
故选C.
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