（x）-2x·x-x-2x+1=0求x的值,

[(x^2)^2-2x^2*x+x^2]-(2x^2+2x+1)=0 即(x^2-x)^2-(2x^2-2x)-4x+1=0 [(x(x-1))^2-2(x(x-1))+1]-4x=0 (x(x-1)-1)^2-4x=0 上式成立必须x>=0 x(x-1)-1=2x^0.5 x^2=x+2x^0.5+1=(x^0.5+1)^2 x=x^0.5+1 令x^0.5=y 即y^2=y+1 解y1=(1+5^0.5)/2 ,y2=(1-5^0.5)/2 x1=[(1+5^0.5)/2]^2 ,x2=[(1-5^0.5)/2]^2