【答案】
分析:解法一:利用分步乘法原理展開式中的常數(shù)項(xiàng)是三種情況的和,
解法二:先將
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/0.png)
利用完全平方公式化成二項(xiàng)式,利用二項(xiàng)展開式的通項(xiàng)公式求得第r+1項(xiàng),令x的指數(shù)為0得常數(shù)項(xiàng).
解答:解法一:(|x|+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/1.png)
-2)
3=(|x|+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/2.png)
-2)(|x|+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/3.png)
-2)(|x|+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/4.png)
-2)得到常數(shù)項(xiàng)的情況有:
①三個(gè)括號(hào)中全取-2,得(-2)
3;
②一個(gè)括號(hào)取|x|,一個(gè)括號(hào)取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/5.png)
,一個(gè)括號(hào)取-2,得C
31C
21(-2)=-12,
∴常數(shù)項(xiàng)為(-2)
3+(-12)=-20.
解法二:(|x|+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/6.png)
-2)
3=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/7.png)
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/8.png)
)
6.
設(shè)第r+1項(xiàng)為常數(shù)項(xiàng),
則T
r+1=C
6r•(-1)
r•(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212803714136863/SYS201310232128037141368009_DA/9.png)
)
r•|x|
6-r=(-1)
6•C
6r•|x|
6-2r,得6-2r=0,r=3.
∴T
3+1=(-1)
3•C
63=-20.
點(diǎn)評(píng):本題考查解決二項(xiàng)展開式的特定項(xiàng)問題的重要工具有二項(xiàng)展開式的通項(xiàng)公式;還有分步乘法原理.