问题 解答题
(1)把 x3+2x2y+y3+2xy2在实数范围内因式分解;
(2)已知x=2009-5
3
,求代数式
x2-2x+1
x2-1
1+
x-3
x+1
+
1
1-
1
1+
1
x
-2(cos30°)2的值.
答案

(1)x3+2x2y+y3+2xy2=(x+y)(x2-xy+y2)+2xy(x+y)=(x+y)(x2+xy+y2);

(2)原式=

(x-1)2
(x+1)(x-1)
×
x+1
2(x-1)
+x+1-2×(
3
2
2=x,

当x=2009-5

3
时,原式=2009-5
3

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