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> (1)解方程:x2-2x-2=0(2)计算:(12)-2-|1-tan60°|+(3-2)0.-数学
(1)解方程:x2-2x-2=0(2)计算:(12)-2-|1-tan60°|+(3-2)0.-数学
题目简介
(1)解方程:x2-2x-2=0(2)计算:(12)-2-|1-tan60°|+(3-2)0.-数学
题目详情
(1)解方程:x
2
-2x-2=0
(2)计算:
(
1
2
)
-2
-|1-tan60°|+(
3
-
2
)
0
.
题型:解答题
难度:中档
来源:不详
答案
(1)由原方程移项,得
x2-2x=2,
等式两边同时加上一次项系数一半的平方1,得
x2-2x+1=2+1,
(x-1)2=3.
∴x1=1+
3
,x2=
1-
3
;
(2)原式=4-|1-
3
|+1
=5-
3
+1
=
6-
3
.
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题目简介
(1)解方程:x2-2x-2=0(2)计算:(12)-2-|1-tan60°|+(3-2)0.-数学
题目详情
(2)计算:(
答案
x2-2x=2,
等式两边同时加上一次项系数一半的平方1,得
x2-2x+1=2+1,
(x-1)2=3.
∴x1=1+
(2)原式=4-|1-
=5-
=6-