已知x=(根号3+根号2)⼀(根号3-根号2),y=(根号3-根号2)⼀(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
2024-12-22 00:48:42
推荐回答(3个)
回答1:

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

=5+2√6

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

[√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

回答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=25-24=1
x+y=5+2√6+5-2√6=10
所以原式=(1+100)/(1-100)
=-101/99

回答3:

xy=1 x+y=10 代数式结果为-101/99