【1】∵-3∈A∴a-3=-3或2a²+5a=-3即a=0或(2a+3)(a+1)=0==>a=-3/2或-1但是,当a=-1时,a-2=-3,2a²+5a=-3∴a=0或-3/2【2】∵a、b∈R又∵{1,a+b,a}={0,a/b,b}所以{1,0}={1,a+b,a}∩{0,a/b,b}∴有a=0或者a+b=0,a/b=1或b=1
