已知x=1/(3的平方根+2的平方根),求x^5+x^4-10x^3-10x^2+2x+1的值

x=1/(√3 +√2)= √3 -√2
x^2=5 -2 √6
x^4=49 -20 √6
x^3=(√3 -√2)(5 -2 √6) =9 √3 -11 √2
x^5=(√3 -√2)(49 -20 √6)=89 √3 - 109 √2
x^5+x^4-10x^3-10x^2+2x+1 = 89 √3 - 109 √2 + 49 -20 √6
- 90 √3 + 110 √2 - 50 + 20 √6 + 2√3 - 2√2 +1 = √3 -√2

x=1/(3^0.5+2^0.5)(上下同时乘以3^0.5-2^0.5)=3^0.5-2^0.5
原式=x^5-10x^3+x^4-10x^2+x+1+x
=x^3(x^2-10)+x^2(x^2-10)+x+1+x
=(x+1)(x^4-10x^2)+(x+1)+x
=(x+1)(x^4-10x^2+1)+x
=(x+1)((x^2-5)^2-24)+x
（因为x^2=5-2*6^0.5所以(x^2-5)^2=24）
=x
=√3-√2

分母有理化 x = 3^0.5-2^0.5
原式 = x^5+x^4-10x^3-10x^2+2x+2-1
=(x+1)(x^4-10x^2+2)-1
=3^0.5-2^0.5