Question

\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)

Answer 1

\(\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\)

<=> \(\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\)

<=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)<=>\(\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)

Vậy x\(\in\){4,5,6}

<=>

Answer 2

\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)

\(\Rightarrow\left(x-5\right)^{2016}\left[1-\left(x-5\right)^2\right]=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{2016}=0\\\left(x-5\right)^2=1-0=1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\\x-5=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)

Answer 3

 x=5 hoặc x=6

Answer 4

\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)

\(\Rightarrow\left(x-5\right)^{2016}-\left(x-5\right)^{2018}=0\)

\(\Rightarrow\left(x-5\right)^{2016}\left[1-\left(x-5\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{2016}=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{2016}=0\\\left(x-5\right)^2=1\end{cases}}\)

TH 1 : \(\left(x-5\right)^{2016}=0\Rightarrow x-5=0\Rightarrow x=5\)

TH 2 : \(\left(x-5\right)^2=1\Rightarrow\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

Vậy  \(x\in\left\{5;6;4\right\}\)

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