Bạn tham khảo
`ĐK:{(x>=0; y>0),(4y^2+2y-2x+4>=0):}`
`{(log_4 (x^2+y^2)log_4 (2x)+1=log_4 (x+3)),(\log_4 (xy+1)-log_4 (4y^2+2y-2x+4)=log_4 (x/y)-1):}`
`<=>{(4(x^2+y^2)/(2x)=x+3y),((xy+1)/(4y^2+2y-2x+4)=x/(4y)):}`
`<=>{(x^2-3xy+2y^2=0),(x^2-xy-2x+2y=0):}`
`<=>{((x-y)(x-2y)=0),((x-y)(x-2)=0):}`
`<=>{([(x=y)(x=2y):}),([(x=y),(x=2):}):}`
Vậy, `S={1;2}`
`\overline{xyxyxyxy}=\overline{12121212}`