Vì \(\left(x+2y-3\right)^{2016}\ge0;\left|2x+3y-5\right|\ge0\forall x;y\)
\(\Rightarrow\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|\ge0\forall x;y\)
Mà \(\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|=0\) \(\Leftrightarrow\left(x+2y-3\right)^{2016}=0\) ; \(\left|2x+3y-5\right|=0\)
\(\Rightarrow x+2y-3=0;2x+3y-5=0\)
\(\Leftrightarrow x+2y=3;2x+3y=5\)
\(\Rightarrow x=3-2y\)
\(\Rightarrow2\left(3-2y\right)+3y=5\Leftrightarrow6-4y+3y=5\Leftrightarrow6-y=5\Rightarrow y=1\)
\(\Rightarrow x=3-2.1=1\)
Vậy \(x=1;y=1\)