解A={-1,1},由B包含于A,B≠空集,故B={-1}或B={1}或B={-1,1},如果B={-1},则-1是x^2-2ax+b=0的惟一解,故2次方程的判别式(-2a)^2-4b=0,a^2=b,且(-1)^2+2a+b=0,2a+b+1=0,解关于a,b方程组a^2=b,2a+b+1=0,解得a=-1,b=1；如...