`a+b+c=1/a +1/b +1/c =1` tính `P=a^(2019) + b^(2019) + c^(2019) `

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1. `1/a+1/b+1/c=1=1/(a+b+c)` vì` a+b+c=1`

   `⇔1/a+1/b+1/c-1/(a+b+c)=0`

   `⇔(a+b)/(ab) +(a+b)/[c(a+b+c)]=0`

   `⇔(a+b).(1/(ab)+1/[c(a+b+c)]=0`

   `⇔(a+b).(ca+cb+c^2+ab)/(abc)=0`

   `⇒(a+b)(c+a)(b+c)=0`

   ⇒\(\left[ \begin{array}{l}a+b=0\\c+a=0\\b+c=0\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}a=-b\\c=-a\\b=-c\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}c=1\\b=1\\c=1\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}p=1\\p=1\\p=1\end{array} \right.\) 

   `⇒p=1`

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