若3x^2—x=1,则9x^4+12x^3-2x^2-7x+2000=?

由已知3x²-x=1得3x² = x +1,多次代入原式以降次得
原式=9x^4+12x³-3x²-7x+2000
=(3x²)²+ (3x²)*4x -3x²-7x+2000
=(x +1)²+ (x +1)*4x -(x +1)-7x+2000
=5x²-2x+2000
=(5/3)*3x²-2x+2000
=(5/3)* (x +1)-2x+2000
=[(5- x )/3]+2000
再由已知3x²-x=1得3x² - x -1=0,解方程得x=(1±√13)/6,代入上面化简后的结果得
原式=(1/3)*[ 5-(1±√13)/6 ]+2000=[ (29±√13)/18 ]+2000

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!