(1)计算:275-327+12;(2)计算:2sin60°-cot30°+tan45°.(3)解方程:x2-6x+1=0.-数学

题目简介

(1)计算:275-327+12;(2)计算:2sin60°-cot30°+tan45°.(3)解方程:x2-6x+1=0.-数学

题目详情

(1)计算:2
75
-3
27
+
12

(2)计算:2sin60°-cot30°+tan45°.
(3)解方程:x2-6x+1=0.
题型:解答题难度:中档来源:不详

答案

(1)原式=10
3
-9
3
+2
3
=3
3


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

(3)原方程可转化为(x-3)2=8
即x-3=±2
2

∴x1=3+2
2
,x2=3-2
2

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