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ab+ac+bc=0.5[(a+b+c)²-(a²+b²+c²)]=0所以ab=-(a+b)c=-(1-c)c又因為a+b=1-c所以a、b是方程x²-(1-c)x-(1-c)c=0兩異實根△=(1-c)²+4(1-c)c=(1-c)(1+3c)>0,所以-1/3<c...
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