`A=1/x+1/y-1/(x^2y^2)`
`=(x+y)/(xy)-1/(x^2y^2)`
`=2/(xy)-1/(x^2y^2)`
`=-(1/(x^2y^2)-2/(xy)+1)+1`
`=-(1/(xy)-1)^2+1<=1`
Dấu "=" `<=>1/(xy)=1<=>xy=1`
`x+y=2=>x=2-y`
`=>xy=1`
`<=>2y-y^2=1`
`=>y^2-2y+1=0`
`<=>(y-1)^2=0`
`<=>y=1<=>x=1`
Vậy `Max_A=1<=>x=y=1`