题目内容
(1)计算:
①2
-
-(
+
-2
);
②(3
-2
+
)÷2
.
(2)解方程:
①3x2+8x-3=0;
②x-2=x(x-2);
③3x2-4x-2=0(配方法).
①2
|
|
| 18 |
| 2 |
|
②(3
| 12 |
|
| 48 |
| 3 |
(2)解方程:
①3x2+8x-3=0;
②x-2=x(x-2);
③3x2-4x-2=0(配方法).
(1)①原式=2×
-
-(3
+
-
)
=
-
-3
-
+
=-4
+
;
②原式=
-
+
=
×2-
+
×4
=3-
+2
=4
;
(2)①(3x-1)(x+3)=0,
∴3x-1=0或x+3=0,
∴x1=
,x2=-3;
②(x-2)-x(x-2)=0,
(x-2)(1-x)=0,
∴x-2=0或1-x=0,
∴x1=2,x2=1;
③3x2-4x=2,
变形得:x2-
x=
,
配方得:x2-
x+
=
+
,即(x-
)2=
,
开方得:x-
=±
,
则x1=
,x2=
.
| ||
| 4 |
| ||
| 2 |
| 2 |
| 2 |
| 2 |
| 3 |
| 3 |
=
| ||
| 2 |
| ||
| 2 |
| 2 |
| 2 |
| 2 |
| 3 |
| 3 |
=-4
| 2 |
| 2 |
| 3 |
| 3 |
②原式=
| 3 |
| 2 |
| 4 |
|
| 1 |
| 2 |
| 16 |
=
| 3 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
=3-
| 1 |
| 3 |
=4
| 2 |
| 3 |
(2)①(3x-1)(x+3)=0,
∴3x-1=0或x+3=0,
∴x1=
| 1 |
| 3 |
②(x-2)-x(x-2)=0,
(x-2)(1-x)=0,
∴x-2=0或1-x=0,
∴x1=2,x2=1;
③3x2-4x=2,
变形得:x2-
| 4 |
| 3 |
| 2 |
| 3 |
配方得:x2-
| 4 |
| 3 |
| 4 |
| 9 |
| 2 |
| 3 |
| 4 |
| 9 |
| 2 |
| 3 |
| 10 |
| 9 |
开方得:x-
| 2 |
| 3 |
| ||
| 3 |
则x1=
2+
| ||
| 3 |
2-
| ||
| 3 |
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