A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

=\(\dfrac{1}{3}-\dfrac{1}{2017}\)

= \(\dfrac{2014}{6051}\)

\(\dfrac{x}{5}=\dfrac{2}{3}\)

\(\Rightarrow x.3=2.5\)

\(\Rightarrow x=\dfrac{2.5}{3}\)

\(\Rightarrow x=\dfrac{10}{3}\)

Vậy x=\(\dfrac{10}{3}\)

(x-2).(x+4)=0

*TH1:x-2=0

x=2

*TH2: x+4 =0

x= -4

Vậy x\(\in\) {-4;2}