Question

已知x=(根号3+根号2)/(根号3-根号2)，y=(根号3-根号2)/(根号3+根号2)，求代数式

(根号xy+(x+y)^2)/(根号xy-(x+y)^2)的值。要过程

Answer 1

x=(√3+√2)/(√3-√2)
=(√3+√2)^2/[(√3-√2)(√3+√2)]

=5+2√6

y=(√3-√锋庆2)/(√3+√2)
=(√3-√2)^2/[(√3-√2)(√3+√蠢基游2)]
=5-2√6

[√带销xy+(x+y)^2]/[√xy-(x+y)^2]

={√[(5+2√6)(5-2√6)]+(5+2√6+5-2√6)^2}/{√[(5+2√6)(5-2√6)]-(5+2√6+5-2√6)^2}
=[1+10^2]/[1-10^2]
=101/(-99)
=-101/99

Answer 2

x=(√3+√2)/(√3-√2)=(√3+√2)²/(3-2)=5+2√6
y=(√3-√2)/(√3+√2)=(√3-√2)²/(3-2)=5-2√6
那么xy=5²-(2√6)²掘并庆销=25-24=1
x+y=5+2√6+5-2√判差迹6=10
所以原式=(1+100)/(1-100)
=-101/99

Answer 3

xy=1 x+y=10 代数式结果为-101/99