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分式化简(a^2-a+1)/(a^2+a+1)+2a(a-1)

2008-07-18 11:55:24有***
(a^2-a+1)/(a^2+a+1) + [2a(a-1)(a-1)]/(a^4+a^2+1) + [2a^2(a^2-1)(a^2-1)]/(a^8+a^4+1)分式化简(a^2-a+1)/(a^2+a+1)+[2a(a-1)(a-1)]/(a^4+a^2+1)+[2a^2(a^2-1)(a^2-1)]/(a^8+a^4?

最佳回答

  •   a^4+a^2+1=a^4+2a^2+1-a^2=(a^2+1)^2-a^2=(a^2+a+1)(a^2-a+1), a^8+a^4+1=(a^4+a^2+1)(a^4-a^2+1) 因此(a^2-a+1)/(a^2+a+1) + [2a(a-1)(a-1)]/(a^4+a^2+1) =1-2a/(a^2+a+1)+2a(a-1)(a-1)]/(a^4+a^2+1) =1-2a(a^2-a+1)/[(a^2+a+1)(a^2-a+1)] + [2a(a-1)(a-1)]/(a^4+a^2+1) =1-2a[(a^2-a+1)-(a-1)^2]/[a^4+a+1]=1-2a^2/[a^4+a^2+1] 因此(a^2-a+1)/(a^2+a+1) + [2a(a-1)(a-1)]/(a^4+a^2+1) + [2a^2(a^2-1)(a^2-1)]/(a^8+a^4+1)=1-2a^2/[a^4+a^2+1] +[2a^2(a^2-1)(a^2-1)]/(a^8+a^4+1)=1-2a^2*[(a^4-a^2+1)-(a^2-1)^2]/[a^8+a^4+1]=1-2a^4/[a^8+a^4+1]=[a^8-a^4+1]/[a^8+a^4+1]。
      
    2008-07-18 12:50:54
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