Mình chỉ trả lời 1 câu thôi:
(2x+1)2018=(2x+1)2016x(2x+1)2
=> Vì (2x+1)2016 = (2x+1)2018
=> (2x+1)2018 - (2x+1)2016 = 0
=> (2x+1)2016x(2x+1)2 - (2x+1)2016 =0
=> (2x+1)2016 x ((2x+1)2 - 1)=0
=> TH1: TH2:
=> (2x+1)2016 =0 (2x+1)2 - 1 =0
=> 2x+1=0 => (2x+1)2 = 1
=> 2x=-1 => ThA: 2x+1=1 ThB: 2x+1 = -1
=>x=-1/2 => x=0 => x=-1
Vậy x thuôc tập hợp:-1/2:0:-1
Ta có : \(\left(2x+1\right)^{2016}=\left(2x+1\right)^{2018}\)
\(\Rightarrow\left(2x+1\right)^{2016}-\left(2x+1\right)^{2018}=0\)
\(\Rightarrow\left(2x+1\right)^{2016}\left[1+\left(2x+1\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\1+\left(2x+1\right)^2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x+1=\hept{\begin{cases}1\\-1\end{cases}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\hept{\begin{cases}0\\-1\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x+1=\orbr{\begin{cases}1\\-1\end{cases}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\hept{\begin{cases}0\\-1\end{cases}}\end{cases}}\)
