来自刘士喜的问题
(2012•随州)设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则(ab2+b2−3a+1a)5=______.
(2012•随州)设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则(ab2+b2−3a+1a)5=______.
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2020-05-01 05:39
(2012•随州)设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则(ab2+b2−3a+1a)5=______.
(2012•随州)设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则(ab2+b2−3a+1a)5=______.
∵a2+2a-1=0,b4-2b2-1=0,∴(a2+2a-1)-(b4-2b2-1)=0,化简之后得到:(a+b2)(a-b2+2)=0,若a-b2+2=0,即b2=a+2,则1-ab2=1-a(a+2)=1-a2-2a=-(a2+2a-1),∵a2+2a-1=0,∴-(a2+2a-1)=0,与题设矛盾∴a-b...