$解:由x²+4y²-8x-4y+17=0,\ $
$得(x-4)²+(2y-1)²=0,\ $
$所以x-4=0,2y-1=0,解得x=4,y=\frac{1}{2}.\ $
$因为\frac{x²y-4y^{3} }{x²+4xy+4y²} ·(\frac {4xy}{x-2y}+x)\ $
$=\frac{y(x+2y)(x-2y)}{(x+2y)²} ·\frac{x(x+2y)}{(x-2y)}\ $
$=xy,$
$所以原式=4×\frac{1}{2}=2.$