A={x/y=根号(2x-x^2)}2x-x²≥0即x²-2x≤0解得A=[0,2] B={y/y=2^x/(2^x-1)(x>0)} y=[(2^x-1)+1]/(2^x-1)=1+1/(2^x-1)∵x>0∴2^x>1∴1/(2^x-1)>0∴B=(1,+∞)∴AUB=[0,+∞)A∩B=(1,2]∴A×B={x|X∈A...