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令f(x)=(x2-x+1)-(-2m2-2mx)=x2+(2m-1)x+(2m2+1),判別式為△=(2m-1)2-4(2m2+1)=-4m2-4m-3.令g(m)=-4m2-4m-3.判別式為△′=(-4)2-4×(-4)×(-3)=-32<0,∴g(m)<0恒成立.∴f(x)>0恒…
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