若3x²-x=1,求代数式6x³+7x²-5x+2008的值

t=[3 -1 -1]
x=roots(t)
t =
3 -1 -1

x =
0.7676
-0.4343
y=6.*x.^3+7*x.^2-5.*x+2008
y =
2011
2011

方法二：
由：
3x^2-x=1 得：
6x^2-2x=2;
6x^3-2x^2=2x;
x^2-x=1-2x^2;
所以：
6x^3+7x^2-5x+2008
=6x^3+(6x^2-2x)+x^2-3x+2008
=6x^3+(6x^2-2x)+(x^2-x)-2x+2008
=6x^3+2+(1-2x^2)-2x+2008
=(6x^3-2x^2)-2x+2+1+2008
=2x-2x+2011
=2011

6x³+7x²-5x+2008
=6x³+13x²-5x-6x²+2008
=x[(6x²+13x -5)-6x]+2008
=x[(2x+5)(3x-1)-6x]+2008
=x(2x+5)(3x-1) -6x² +2008 由题意得 x(3x-1)=1
=2x+5 -6x² +2008
=-2(3x²-x)+5+2008
=-2+5+2008
=2011

原式=6x³-2x²+9x²-3x-2x+2008
=2x(3x²-x)+3(3x²-x)-2x+2008
=2x+3+2x-2008
=2011