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  • 若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值

    若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值
    數(shù)學(xué)人氣:105 ℃時(shí)間:2020-06-12 03:36:33
    優(yōu)質(zhì)解答
    a^2+3a+1=0 => a^3+3a^2+a=0;a^4+3a^3+a^2=0;2a^2+6a+2=0
    =(a^4-3a^2+1-a^4-3a^3-a^2)/(a^3+5a^2+1)
    =(-3a^3-4a^2+1)/(a^3+5a^2+1)
    =-3+(11a^2+4)/(a^3+5a^2+1)
    =-3+(11a^2+4)/(a^3+5a^2+1-a^3-3a^2-a)
    =-3+(11a^2+4)/(2a^2-a+1)
    =-3+(11a^2+4)/(2a^2-a+1-a2^2-6a-2)
    =-3+(11a^2+4)/(-7a-1)
    =-3+(33a+7)/(7a+1)
    =-3+4+(5a+3)/(7a+1)
    =1+(5a+3)/(7a+1)
    又a=(-3(+ -)genhao(5))/2
    帶入得=1+(-9(+ -) 5genhao(5))/(-19(+-)7genhao(5))
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