已知x^2-4x+1=0;求(1)x^4-2x^3-8x^2+6x+2012 (2)5x^2-12x+2/x^2

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  • 已知x^2-4x+1=0;

    x^2=4x-1

    (1)x^4-2x^3-8x^2+6x+2012

    =x^4-4x^3+x^2+2x^3-9x^2+6x+2012

    =x^2(x^2-4x+1)+2(x^3-4x^2+x)-x^2+4x+2012

    =x^2(x^2-4x+1)+2x(x^2-4x+1)-(x^2-4x+1)+1+2012

    =2013

    (2)5x^2-12x+2/x^2

    =((3x^2-12x+3)+2x^2-1)/x^2

    =(3(x^2-4x+1)+2x^2-1)/x^2

    =(2x^2-1)/x^2

    =2-1/x^2

    x^2-4x+1=0

    x=4+√16-4/2=2+√3

    或x=4-√16-4/2=2-√3

    当x=2+√3

    则上式==2-1/x^2

    =2-1/(2+√3)^2

    =2-(2-√3)^2/(2+√3)^2(2-√3)^2

    =2-(4-2√3+3)/(4-3)^2

    =2√3-5

    当x=2-√3

    则上式==2-1/x^2

    =2-1/(2-√3)^2

    =2-(2+√3)^2/(2+√3)^2(2-√3)^2

    =2-(4+2√3+3)/(4-3)^2

    =-2√3-5