
(1)计算:12+|?3|?2tan60°+(?1+2)0;(2)化简:(a2a?2?4a?2)?1a2+2a
(1)计算:12+|?3|?2tan60°+(?1+2)0;(2)化简:(a2a?2?4a?2)?1a2+2a....
(1)计算:12+|?3|?2tan60°+(?1+2)0;(2)化简:(a2a?2?4a?2)?1a2+2a.
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(1)原式=2
+3-2×
+1
=4;
(2)原式=
?
=
?
=
.
故答案为:4;
.
3 |
3 |
=4;
(2)原式=
a2?4 |
a?2 |
1 |
a2+2a |
=
(a+2)(a?2) |
a?2 |
1 |
a(a+2) |
=
1 |
a |
故答案为:4;
1 |
a |
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