已知x^2+4y^2=4xy-|x+y-9|,则(y-2)^x-4 -(x-5)^y-1=??

2024-12-23 09:52:25
推荐回答(2个)
回答1:

x^2+4y^2=4xy-|x+y-9|
x^2+4y^2-4xy+|x+y-9|=0
(x-2y)^2+|x+y-9|=0
平方和绝对值都大于等于0
相加等于0,若有一个大于0,则另一个小于0,不成立。
所以两个都等于0
所以x-2y=0
x+y-9=0
所以x=6,y=3

(y-2)^x-4 -(x-5)^y-1
=1^2-1^2
=0

回答2:

x^2+4y^2=4xy-|x+y-9|,
(x^2+4y^2-4xy)+|x+y-9|=0
(x-2y)^2+|x+y-9|=0
(x-2y)^2=0,|x+y-9|=0
x-2y=0,x+y-9=0

x-2y=0........(1)
x+y-9=0.......(2)
(2)-(1),得
3y-9=0
3y=9
y=3
代入(1),得
x-2*3=0
x-6=0
x=6

(y-2)^x-4 -(x-5)^y-1
=(3-2)^(6-4)-(6-5)^(3-1)
=1^2-1^2
=1-1
=0