Question

已知x>y且x+y=5 xy=4求根号x-根号y分之根号x+根号y

Answer 1

已知x>y且x+y=5 xy=4
(x-y)^2=(x+y)^2-4xy=5^2-4*4=9
x-y=3
∴根号x-根号y分之根号x+根号y
=(根号x+根号y)(根号x+根号y)/[(根号x-根号y)(根号x+根号y)]
=[x+y+2√(xy)]/(x-y)
=[5+2√4]/3
=(5+4)/3
=3
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Answer 2

x+y=5
xy=4所以y=4/x
所以x+4/x=5
解得x=4或1
因为x>Y
所以X=4Y=1
带入得原式=3

Answer 3

结果是3
由 xy=4得x与y的关系为y=4/x，代人x+y=5得：x·x-5x+4=0，解得X1=1，y1=4或X2=4，y2=1
因为x>y，所以x=4，y=1.代人根号x-根号y分之根号x+根号y=3/1=3

Answer 4

答案是3 x+y=5 xy=4解得x=1 y=4或x=4 y=1 已知x>y 所以x=4 y=1 带入根号x-根号y分之根号x+根号y

Answer 5

(根号x+根号y)^2/(根号x-根号y）^2
=(x+y+2根号xy)/(x-y+2根号xy)
=(5+4)/(5-4)=9
所以：根号x-根号y分之根号x+根号y=3

Answer 6

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过程