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已知实数a满足a的平方- a-1=0,求a的8次方+7a的负四次方
人气:134 ℃ 时间:2019-12-18 05:13:11
解答
由a^2-a-1=0得,a^2=a+1,于是a^4=(a+1)^2=a^2+2a+1=3a+2 a^8=(3a+2)^2=9a^2+12a+4=21a+13 a^8+7a^-4=21a+13+7/(3a+2) =(63a^2+81a+26+7)/(3a+2) =(63a+63+81a+33)/(3a+2) =(144a+96)/(3a+2)=48
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