题目内容
(1)计算:|-1|+
+(-3.14)0-(
)-1.
(2)解方程:2x2-6x-1=0.
(3)如图,在平行四边形ABCD中,E为BC中点,AE和延长线与DC的延长线相交于点F.证明:△ABE≌△FCE.
1 |
2 |
8 |
1 |
2 |
(2)解方程:2x2-6x-1=0.
(3)如图,在平行四边形ABCD中,E为BC中点,AE和延长线与DC的延长线相交于点F.证明:△ABE≌△FCE.
(1)原式=1+
+1-2
=
;
(2)∵2x2-6x-1=0,
∴x2-3x=
,
∴x2-3x+
=
+
.
∴(x-
)2=
,
∴x-
=±
,
∴x1=
,x2=
;
(3)证明:∵四边形ABCD是平行边形,
∴AB∥CD,
∴∠EAB=∠F,
∵E是BC中点,
∴BE=CE,
∵在△ABE和△FCE中,
,
∴△ABE≌△FCE(AAS).
2 |
=
2 |
(2)∵2x2-6x-1=0,
∴x2-3x=
1 |
2 |
∴x2-3x+
9 |
4 |
1 |
2 |
9 |
4 |
∴(x-
3 |
2 |
11 |
4 |
∴x-
3 |
2 |
| ||
2 |
∴x1=
3+
| ||
2 |
3-
| ||
2 |
(3)证明:∵四边形ABCD是平行边形,
∴AB∥CD,
∴∠EAB=∠F,
∵E是BC中点,
∴BE=CE,
∵在△ABE和△FCE中,
|
∴△ABE≌△FCE(AAS).
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