第一题:已知y^2+3y-1=0,求y^4/(y^8-3y^4+1)的值.

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  • 第一题:已知y^2+3y-1=0,求y^4/(y^8-3y^4+1)的值.

    y^2+3y-1=0,两边除以y,得 y+3-1/y = 0 ,y – 1/y = -3

    两边 平方,y ^2 + 1/y^2 -2 = 9,y ^2 + 1/y^2 = 11,

    再次平方,y^4 + 1/y^4 + 2 = 121,y^4 + 1/y^4 = 119,

    y^4/(y^8-3y^4+1) = 1/(y^4 – 3 + 1/y^4) = 1/(119 - 3) = 1 / 116

    第二题:计算(2007^3-2*2007^2+1)/(2007^3+2007^2-3*2007-2)

    (x^3 – 2x^2 + 1 ) / (x^3 + x^2 – 3x – 2)

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

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

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

    (2007^3-2*2007^2+1)/(2007^3+2007^2-3*2007-2)

    = (2007-1)/(2007+2)

    = 2006/2009