x^2+x-1=0两边乘以x,得到x^3+x^2-x=0
x^3+x^2-x=0减去x^2+x-1=0可以得到x^3+x^2-x-(x^2+x-1)=0
即得到,x^3+x^2-x-x^2-x+1=0,所以x^3-2x+1=0
x^3-2x+4=x^3-2x+1+3=3
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x^3-2x+4=x(x^2+x-1)-x^2-x+4
=x(x^2+x-1)-(x^2+x-1)+3
由已知条件x^2+x-1=0
所以x(x^2+x-1)-(x^2+x-1)+3=3
答案是3
即x²+x=1
所以原式=x³+x²-x²-2x+4
=x(x²+x)-x²-2x+4
=x-x²-2x+4
=-(x²+x)+4
=-1+4
=3
答:
x^2+x-1=0
x^2=1-x
x^3-2x+4
=(1-x)x-2x+4
=x-x^2-2x+4
=-x^2-x+4
=-1+4
=3