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
+,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
+, a+b khác 0 chia cả 2 vế cho a+b ta được:
1/ab=-1/(a+b+c)c
<=> -ab=ac+bc+c^2
<=> ab+ac+bc+c^2 =0
<=> (a+c)(b+c)=0
<=> hoặc: a+c=0 hoặc: b+c =0
+, a+c =0 =>1/a^2009 + 1/b^2009 + 1/c^2009 = 1/a^2009+b^2009+c^2009=1/b^2019
+, b+c=0 =>1/a^2009 + 1/b^2009 + 1/c^2009 = 1/a^2009+b^2009+c^2009=1/a^2019
=> điều phải chứng minh
