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

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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
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

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?