问题 解答题
计算:
(1)(-2)2×
12
-4
3
(4-
3
)+
8
2-
3

(2)已知x=
1
3
,求
x2 -2x+1
x2-x
-
1-2x+x2
x-1
的值.
(3)2sin30°+cos60°-tan60°-tan30°+cos245°.
答案

(1)原式=4×2

3
-16
3
+12+8(2+
3
)=8
3
-16
3
+12+16+8
3
=28;

(2)∵x=

1
3
<1,

∴原式=

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

(3)原式=2×

1
2
+
1
2
-
3
×
3
3
+(
2
2
2=1.

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