Question

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