`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`
