`1/a+1/b+1/c=1/(a+b+c)`
`<=> 1/a+1/b= 1/(a+b+c)-1/c`
`<=> (a+b)/[ab]=[-(a+b)]/[(a+b+c)c]`
- Nếu `a+b=0`
`=> a=-b`
`=> 1/[a^2019] + 1/[b^2019] + 1/[c^2019] = 1/c^2019; 1/(a^2019+b^2019+c^2019)=1/[c^2019]`
`=> 1/[a^2009] + 1/[b^2009] + 1/[c^2009] = 1/(a^2009+b^2009+c^2009)`
- Nếu `a+b \ne 0`
`⇒ 1/[ab]=-1/[(a+b+c)c]`
`<=> -ab=ac+bc+c^2`
`<=> ab+ac+bc+c^2 =0 `
`<=> (a+c)(b+c)=0`
`<=>` \(\left[ \begin{array}{l}a+c=0\\b+c=0\end{array} \right.\)
- Nếu `a+c =0`
`=>1/[a^2009] + 1/[b^2009] + 1/[c^2009] = 1/[a^2009+b^2009+c^2009]=1/[b^2019]`
- Nếu `b+c=0`
`=>1/[a^2009] + 1/[b^2009] + 1/[c^2009] = 1/[a^2009+b^2009+c^2009]=1/[a^2019]`
Xin hay nhất !
