构造方程x^2-2√3x+1=0,其两根为x1=√3-√2,x2=√3+√2,
原题等价于x1=√3-√2的取值，
则原式=x^5+x^4-10x^3-10x^2+2x+1
=x^4*(x+1)-10x^2*(x+1)+(x+1)+x
=[(x^2+1)^2-12x^2]*(x+1)+x
=(x^2-2√3x+1)*(x^2+2√3x+1)*(x+1)++x----(平方差公式展开)
=0+x
=x
=√3-√2
x5+x4-10x3-10x2+2x+1
=x4(x+1)-10x2(x+1)+2x+1
=(x+1)(x4-10X2)+2x+1
=(x+1)(x2)(x2-10)+2x+1
=(x+1)(5-2根号6)(5-2根号6-10)+2x+1
=-(x+1)(5-2根号6)(5+2根号6)+2x+1
=-(x+1)+2x+1
=-x-1+2x+1
=x
x=√3-√2
x^5+x^4-10x^3-10x^2+2x+1
=x^2(x^2-10)(x+1)+2x+1
=(5-2√6)(-5-2√6)(√3-√2+1)+2(√3-√2)+1
=√3-√2
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