(1)计算:(12)-1+8+|1-2|0-2sin60°-tan60°.(2)解方程:2(x+1)2x2+x+1x-6=0.-数学

题目简介

(1)计算:(12)-1+8+|1-2|0-2sin60°-tan60°.(2)解方程:2(x+1)2x2+x+1x-6=0.-数学

题目详情

(1)计算:(
1
2
)-1+
8
+|1-
2
|0-2sin60°-tan60°

(2)解方程:
2(x+1)2
x2
+
x+1
x
-6=0
题型:解答题难度:中档来源:不详

答案

(1)原式=2+2
2
+1-2×
3
2
×
3

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



(2)设class="stub"x+1
x
=y,则原方程转化为2y2+y-6=0,
解得,y=class="stub"3
2
或y=-2,
当y=class="stub"3
2
时,class="stub"x+1
x
=class="stub"3
2
,解得x=2;
当y=-2时,class="stub"x+1
x
=-2,解得x=-class="stub"1
3

经检验,x1=2,x2=-class="stub"1
3
是原方程的解.
故答案为:2
2
;x1=2,x2=-class="stub"1
3

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