问题
解答题
计算: (1)6tan30°-
(2)
|
答案
(1)原式=6×
-3 3
×3
-2×3 2 2 2
=2
-3
-3 2
;2
(2)原式=
×3
-2-3 2 1 (
)21 3
=
-2-93 2
=-
.19 2
计算: (1)6tan30°-
(2)
|
(1)原式=6×
-3 3
×3
-2×3 2 2 2
=2
-3
-3 2
;2
(2)原式=
×3
-2-3 2 1 (
)21 3
=
-2-93 2
=-
.19 2