已知,a2+3a-1=0,b4-3b2-1=0,且1-ab2≠0,则(ab2+b2+1a)5的值为_.
问题描述:
已知,a2+3a-1=0,b4-3b2-1=0,且1-ab2≠0,则(
)5的值为______. ab2+b2+1 a
答
将a2+3a-1=0,b4-3b2-1=0两式相减得:a2-b4+3a+3b2=0,a2-b4+3a+3b2=(a2-b4)+(3a+3b2)=(a+b2)(a-b2)+3(a+b2)=(a+b2)(a-b2+3)=0,若a-b2+3=0,则1-ab2=1-a(a+3)=-(a2+3a-1)=0,而已知1-ab2≠0,所...