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  • 21.當(dāng)n為自然數(shù)時,有x^6n+1/x^6n=2 ..

    21.當(dāng)n為自然數(shù)時,有x^6n+1/x^6n=2 ..
    21.當(dāng)n為自然數(shù)時,有x^6n+1/x^6n=2
    證明:(1)x+1/x=-1 ==>x^6n+1/x^6n=2
    (2)x+1/x=1 ==>x^6n+1/x^6n=2
    22.證明:(1)a^2+b^2+c^2+d^2-ab-bc-cd-da=0 ==>a=b=c=d
    (2)a^4+b^4+c^4+d^4-4abcd=0 ==>a=b=c=d
    數(shù)學(xué)人氣:575 ℃時間:2020-01-28 23:51:27
    優(yōu)質(zhì)解答
    21.x^2+x+1=0
    x^3-1=(x-1)(x^2+x+1)=(x-1)*0=0
    ==>x^3=1
    ==>x^6n=(x^3)^2n=1^2n=1
    ==>x^6n+1/x^6n=1+1=2
    同理第2個,x^2-x+1=0
    ==>
    x^3+1=(x+1)(x^2-x+1)=(x+1)*0=0
    ==>x^3=-1
    ==>x^6n=(x^3)^2n=(-1)^2n=[(-1)^2]^n=1^n=1
    ==>x^6n+1/x^6n=1+1=2
    22(1)a^2+b^2+c^2+d^2-ab-bc-cd-da=0
    ==>2(a^2+b^2+c^2+d^2-ab-bc-cd-da=0
    ==>(a-b)^2+(a-d)^2+(b-c)^2+(c-d)^2=0
    ==>a=b=c=d
    (2)a^4+b^4+c^4+d^4-4abcd=0
    ==>(a^2-b^2)^2+2a^2*b^2+(c^2-d^2)^2+2c^2*d^2-4abcd=0
    ==>(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2=0
    ==>a^2=b^2;c^2=d^2;ab=cd
    ==>a=b=c=d
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