\(\left(2x-3\right)^{2015}=\left(2x-3\right)^{2013}\)
\(\left(2x-3\right)^{2015}-\left(2x-3\right)^{2013}=0\)
\(\left(2x-3\right)^{2013}\text{[}\left(2x-3\right)^2-1\text{]}=0\)
=> \(\left(2x-3\right)^{2013}=0\) hoặc \(\left(2x-3\right)^2-1=0\)
+) => 2x-3=0=>2x=3=>x=\(\frac{3}{2}\)
+) => \(\left(2x-3\right)^2\)=1
_ 2x-3=1 _2x-3=-1
=>2x=4 =>2x=2
=>x=2 =>x=1
Vậy x={2; 1; \(\frac{3}{2}\)}