ta có \(\frac{1}{a}\)+\(\frac{1}{c}\)=\(\frac{1}{a+b+c}\)-\(\frac{1}{b}\)
⇒\(\frac{a+c}{ac}\)=\(\frac{-\left(a+c\right)}{b\left(a+b+c\right)}\)
⇔\(\left[{}\begin{matrix}a+c=0\\ac=-b\left(a+b+c\right)\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}a=-c\\\left(b+a\right)\left(b+c\right)=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}a=-c\\c=-b\\b=-a\end{matrix}\right.\)
(*) với a=-c ⇒điều cần CM :\(\frac{1}{a^{2019}}\)+\(\frac{1}{b^{2019}}\)+\(\frac{1}{c^{2019}}\)=\(\frac{1}{a^{2019}+b^{2019}+c^{2019}}\)
⇔\(\frac{1}{-c^{2019}}\)+\(\frac{1}{b^{2019}}\)+\(\frac{1}{c^{2019}}\)=\(\frac{1}{-c^{2019}+b^{2019}+c^{2019}}\)
⇔\(\frac{1}{b^{2019}}\)=\(\frac{1}{b^{2019}}\) đúng vậy ta có điều cần CM
tương tự với 2 TH còn lại nhé
