设a^2＋2a－1=0,b^4－2b^2－1=0,且1－ab^2≠0,求［（ab^2＋b^2－3a＋1)／a］^5

a^2+2a-1=0 =>a=-1±√2
b^4-2b^2-1=0 =>b^2=1+√2
∵1-ab^2不等于0
∴a=-1-√2
((ab^2+b^2-3a+1)/a)^5
=[-(√2+1)(1+√2)+1+√2+3(√2+1)+1)/(-1-√2)]^5

有好评？