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三式相加(a+b+c)x²;+(a+b+c)x+(a+b+c)=0 (a+b+c)(x²;+x+1)=0 因x²;+x+1=(x+1/4)²;+3/4≥3/4>0 所以a+b+c=0 故當x=1時,滿足條件 即公共解為x=1
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