Question

已知实数x y满足x的平方+y的平方+四分之五=2x+y，求代数式x+y分之xy=多少 ？？？

Answer 1

解：x∧2-y∧2+5/4＝2x+y可以化为：x∧2-2x+1+y∧2-y+1/4＝0,进一步化为：（x-1）∧2+（y-1/2）∧2＝0，∴x＝1,y＝1/2.∴x+y/xy＝3。祝学习进步*^_^*

Answer 2

x²+y²+5/4=2x+y
解：x²-2x+y²-y+5/4=0
(x-1)²-1+(y-1/2)²-1/4+5/4=0
(x-1)²+(y-1/2)²=0
(x-1)²=0 (y-1/2)²=0
x=1 y=1/2
∴xy/(x+y)=1/3
望采纳！！！！！！

Answer 3

x^2+y^2+5/4=2x+y
(x-1)^2+(y-1/2)^2=0
由非负数性质得
(x-1)^2＝0，(y-1/2)^2=0
x＝1，y＝1/2
xy/(x+y)=1/2/(1+1/2)=1/3

Answer 4

(x-1)^2+(y-1/2)^2=0
x-1=0,y-1/2=0
x=1,y=1/2
代数式x+y分之xy=1/2÷3/2=1/3