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f(x)= x平方+x+a=x(x+1)+a ∵f(m)<0 ∴f(m)=m(m+1)+a<0 即m(m+1)<-a 又∵a>0,且m<m+1 ∴m<0,m+1>0 ∵(m+1)平方≥0 ∴f(m+1)=(m+1)平方+(m+1)+a>0 即f(m+1)>0
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