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證明:(1)充分性:∵m≥2,∴△=m2-4≥0,方程x2+mx+1=0有實根,設x2+mx+1=0的兩根為x1,x2,由韋達定理知:x1x2=1>0,∴x1、x2同號,又∵x1+x2=-m≤-2,∴x1,x2同為負根.(2)必要性:∵x2+mx+1=0的兩個實根x1…
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