题目内容
(1)解方程:(x-3)2+4x(x-3)=0.
(2)解方程:2x2-4x-7=0
(3)计算:
+|
-2|+(2-π)0.
(4)
-
-
(
-1).
(2)解方程:2x2-4x-7=0
(3)计算:
| 12 |
| 3 |
(4)
| 24 |
|
| 2 |
| 3 |
(1)把左边分解因式得:(x-3)(x-3+4x)=0,
x-3=0,x-3+4x=0,
解得:x1=3,x2=
;
(2)a=2,b=-4,c=-7,
b2-4ac=16-4×2×(-7)=72,
x=
=
=
,
x1=
,x2=
;
(3)原式=2
+2-
+1
=
+3;
(4)原式=2
×
-
+
=2
-
+
.
x-3=0,x-3+4x=0,
解得:x1=3,x2=
| 3 |
| 5 |
(2)a=2,b=-4,c=-7,
b2-4ac=16-4×2×(-7)=72,
x=
-b±
| ||
| 2a |
4±
| ||
| 4 |
2±3
| ||
| 2 |
x1=
2+3
| ||
| 2 |
2-3
| ||
| 2 |
(3)原式=2
| 3 |
| 3 |
=
| 3 |
(4)原式=2
| 6 |
| ||
| 2 |
| 6 |
| 2 |
=2
| 3 |
| 6 |
| 2 |
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