x^2-x-1=0,得x^2-x=1 ,x^2-1=x
(x+1)^3+(x^2+x-1)^2-12x=x^3+3x^2+3x+1+(x+x)^2-12x
=x^3+1+3x^2+3x+4x^2-12x
=(x+1)(x^2-x+1)+7x^2-9x
=2(x+1)+7x^2-9x
=7x^2-7x+2
=7(x^2-x)+2
=7+2
=9
我找到类似的:请参考http://zhidao.baidu.com/question/310952742.html?an=0&si=2
(x+1)^3+(x^2+x-1)^2+12x (已知x^2-x-1=0 ,得X²=X+1)
=(x+1)²(x+1)+0²+12x 解方程x²-x-1=0 得x=(1+√5)/2或(1-√5)/2
=(x²+2x+1)(X+1)+12x
=(3x+2)(X+1)+12x
=3x²+5x+2+12x
=3(x+1)+17x+2
=20x+5=15+10√5或15-10√5