Question

1、已知ab^2=-2,那么a和b的值唯一确定吗?请求出ab(-ab^3+a^2b^5-b)的值.

Answer 1

1不唯一确定

ab(-ab^3 + a^2b^5-b)

=-a"b""+ a"'b^6-ab"

=-(ab")"+ (ab")^3-ab"

=-4-8+ 2

=-10

2

(a-b)"*(a b)"

=(a"-b")"

=4"

=16

3

(1) 令x=1

a(5)+a(4)+a(3)+a(2)+a(1)+a(0) =(2*1-1)^5=1

(2)令x=-1

a(0)-a(1)+a(2)-a(3)+a(4)-a(5) =(-1*2-1)^5 = -243

(3)将上面的两个算式相加得到

2(a(0)+a(2)+a(4))

=-242

所以

a(0)+a(2)+a(4) =-121