怎样解设3x2-x=1,求9x4+12x3-3x2-7x+2000的值.

2个回答

  • 3x²-x=1

    x=(1±√13)/6

    9x4+12x³-3x²-7x+2000

    =9x4-3x³+15x³-3x²-7x+2000

    =3x²(3x²-x)+15x³-5x²+2x²-7x+2000

    =3x²(3x²-x)+5x(3x²-x)+2x²-7x+2000

    =3x²+5x+2x²-7x+2000

    =5x²-2x+2000

    =5(3²x-x)/3 - x/3+2000

    =5/3-x/3+2000

    =2001+2/3-(1±√13)/6

    =2001+1/2±(√13)/6

    =2001.5±(√13)/6

    总感觉应该能全部代光而不需要再算x等于多少的,检查了2次都没检查出来,约不掉.结果就这样了.你自己看看吧!