设x=sqrt(a)-1,求x^5+2x^4-ax^3-x^2+(a+1)x-a的值

设x=sqrt(a)-1,求x^5+2x^4-ax^3-x^2+(a+1)x-a的值
设x=sqrt(a)-1,求x^5+2x^4-ax^3-x^2+(a+1)x-a的值

设x=sqrt(a)-1,求x^5+2x^4-ax^3-x^2+(a+1)x-a的值
x=sqrt(a)-1 可得:a=(x+1)^2
原式=x^5+2x^4-(x+1)^2x^3-x^2+[(x+1)^2+1]x-(x+1)^2
=x^5+2x^4-x^5-2x^4-x^3+x^3+2x^2+2x-x^2-2x-1
=x^2-1