.已知x2-2x-1=0,求2x5+4x4-3x3+3x2-x+4的值.

2个回答

  • x²-2x-1=0

    x2-2x+1-2=0

    (x+1)²=2

    x+1=±根号2

    x1=-1+根号2

    x2=-1-根号2

    2x^5+4x^4-3x³+3x²-x+4

    =2x^5+4x^4+x³-4x³-8x²+11x²-4x+3x+4

    =2x³(x+1)²+4x(x+1)²+11x²+3x+4

    =4x³+8x+11x²+3x+4

    =4x³+11x²+11x+4

    =4x³+8x²+3x²+4x+7x+4

    =4x(x+1)²+3x²+7x+4

    =8x+3x²+7x+4

    =3x²+15x+4

    =3x²+6x+9x+3+1

    =3(x+1)²+9x+1

    =6+9x+1

    =9x+7

    当x1=-1+根号2

    原式=-9+9根号2+7=-2+9根号2

    当x2=-1-根号2

    原式=-9-9根号2+7=-2-9根号2