\(\Leftrightarrow\)\(\sum\limits^{2019}_{i=0}\left(x^i\right)=\left(y-x\right)\cdot\sum\limits^{2019}_{i=1}\left(x^{2019-i}y^i\right)\)
\(\Leftrightarrow\left(x+1\right)\cdot\sum\limits^{2018}_{i=0}\left(x^i\right)=\left(y-x\right)\cdot\sum\limits^{2019}_{i=1}\left(x^{2019-i}y^i\right)\)
\(\left(x+1,\sum\limits^{2018}_{i=0}\left(x^i\right)\right)=1\) và \(\sum\limits^{2018}_{i=0}\left(x^i\right)\le\sum\limits^{2019}_{i=1}\left(x^{2019-i}y^i\right)\)(Vì \(x\le y\))
\(\Rightarrow\left\{{}\begin{matrix}x+1=1\\\sum\limits^{2018}_{i=0}\left(x^i\right)=\left(y-x\right)\cdot\sum\limits^{2019}_{i=1}\left(x^{2019-i}y^i\right)\end{matrix}\right.\)\(\Rightarrow x=y=0\)
#Kaito#
