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由题:当x>-2时,f'(x)=[a(x+2)-ax]/[(x+2)^2]=2a/[(x+2)^2]<0,则a<0;
令f(x)=x^2+(m-1)x+1,则f'(x)=2x+m-1=0 => x=(1-m)/2
f(0)=1;
若(1-m)/2<=0,即m>=1时,则f(x)在(0,2]上单增,fmin=f(0)=1>0,无解;
若(1-m)/2>=2,即m<=-3时,则f(x)在(0,2]上单减,fmin=f(2)=2m+3<=0 => m<=-3;
若0<(1-m)/2<2,即-3<m<1时,则f(x)在(0,2]上先减后增,fmin=f[(1-m)/2]=-(m^2)/4+m/2+3/4<=0 => -3<m<=-1;
综上:m<=-1
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