若实数x、y满足x平方+y平方-4x-2y+5=0,求根号8y-2x分之（根号x）+y

因为x^2+y^2-4x-2y+5=0
所以(x-2)^2+(y-1)^2=0
所以x=2 y=1
以下看不懂，你就自己求了

2分之（根号2）+1

x^2+y^2-4x-2y+5=0
(x-2)^2+(y-1)^2=0
x=2，y=1
{根号x +y}/{根号(8y)-2x}
= {根号x +y}/{2根号(2y)-2x}
= {根号x +y}/{2[根号(2y)-x]}
= {根号2 +1}/{2[根号2-2]}
= (根号2 +1)(根号2+2) / {2[2-4]}
= (4+3根号2）/（-4）
= - (4+3根号2) /4

x平方+y平方-4x-2y+5=0
(x-2)^2+(y-1)^2=0
x=2，y=1
根号8y-2x分之（根号x）+y
=[(根号2)+1]/[(根号8) - 4)]
=[3+(2根号2)]/2

x平方+y平方-4x-2y+5=0
(x-2)^2+(y-1)^2=0
(x-2)^2=0
(y-1)^2=0
x=2
y=1
根号8y-2x分之（根号x）+y
=(根号2)+1/根号(8*1-2*2)
=[(根号2)+1]/2
或者=-[(根号2)+1]/2