Ta có:\(\left(x-2\right)^{10}=\left(x-2\right)^{12}\)
\(\Leftrightarrow\left(x-2\right)^{12}-\left(x-2\right)^{10}=0\)
\(\Leftrightarrow\left(x-2\right)^{10}\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^{10}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x\in\left\{3;1\right\}\end{cases}}\)
Vậy \(x\in\left\{-1,2,3\right\}\)
b) Ta có: \(4^{x+2}+4^{x+3}+4^{x+4}+4^{x+5}=85.\left(2^{2016}:2^{2012}\right)\)
\(\Leftrightarrow4^{x+2}\left(1+4+4^2+4^3\right)=85.2^4\)
\(\Leftrightarrow4^{x+2}.85=1360\)
\(\Leftrightarrow4^x=16\)
\(\Leftrightarrow4^x=4^2\)
\(\Leftrightarrow x=2\)
Vậy x=2
