计算:(1)23-1-sin60°+(-25)0-124;(2)[(a+1)(a-2)a2-4a+4-aa2-2a]÷aa-2.-数学

题目简介

计算:(1)23-1-sin60°+(-25)0-124;(2)[(a+1)(a-2)a2-4a+4-aa2-2a]÷aa-2.-数学

题目详情

计算:
(1)
2
3
-1
-sin60°+(-2
5
0-
12
4

(2)[
(a+1)(a-2)
a2-4a+4
-
a
a2-2a
a
a-2
题型:解答题难度:中档来源:不详

答案

(1)class="stub"2
3
-1
-sin60°+(-2
5
)0-
12
4

=
2(
3
+1)
(
3
-1)(
3
+1)
-
3
2
+1-
2
3
4

=
3
+1-
3
2
+1-
3
2

=
3
+2;
(2)原式=[
(a+1)(a-2)
(a-2)2
-class="stub"a
a(a-2)
class="stub"a
a-2

=(class="stub"a+1
a-2
-class="stub"1
a-2
)•class="stub"a-2
a

=class="stub"a
a-2
class="stub"a-2
a

=1.

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