Question

已知x=√3+√2/√3-√2,y=√3-√2/√3+√2,求x^3-xy^2/x^4+2x^3y^2+x^2y^3

Answer 1

x=√3+√2/√3-√2,y=√3-√2/√3+√2
xy=1
x-y=(√3+√2)/(√3-√2)-(√3-√2)/(√3+√2)=[(√3+√2)²-(√3-√2)²]/(3-2)=4√6
x+y==(√3+√2)/(√3-√2)+(√3-√2)/(√3+√2)=[(√3+√2)²+(√3-√2)²]/(3-2)=10
∴x^3-xy^2/x^4y+2x^3y^2+x^2y^3
=x(x²-y²)/[x²y(x²+2xy+y²)]
=(x+y)(x-y)/[xy(x+y)²]
=(x-y)/[xy(x+y)]
=4√6/(1*10)
=2√6/5
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

x=(√3+√2)/(√3-√2),y=(√3-√2)/(√3+√2)
x=5+2√6 y=5-2√6
=>xy=1
=>(x^3-xy^2)/(x^4+2x^3y^2+x^2y^3)=(x^3-y)/(x^4+2x+y)
=[(5+2√6)^3-(5-2√6)]/[(5+2√6)^4+2(5+2√6)+5-2√6]
=(581760+237520√6)/12149

追问

答案对么？

追答

估计你题目抄错啦，不应该这么复杂。

追问

嗯嗯 问的应该是“x^3-xy^2/x^4y+2x^3y^2+x^2y^3”
漏了个y 不好意思。。 再算一遍 。。 O(∩_∩)O谢谢
我给你加分。

追答

xy=1
(x^3-xy^2)/(x^4y+2x^3y^2+x^2y^3)=(x^3-1/x)/(x^3+2x+1/x)
=(x^4-1)/(x^4+2x^2+1)
=(x^2-1)/(x^2+1)
=[(5+2√6)^2-1]/[(5+2√6)^2+1]
=(48+20√6)/(50+20√6)
=(48+20√6)(50-20√6)/100
=2√6/5