△=（a-3)²-4a=a²-10a+9(1)∵x1>0 x2>0∴x1+x2=-a+3>0 x1x2=a>0a²-10a+9≥0∴00 x20 x2-2>0∴(x1-2)+(x2-2)=-a+3-4>0 (x1-2)(x2-2)=x1x2-2(x1+x2)+4=a+2a-6+4>0a²-10a+9≥0∴无解,a的范围是空集...