\(a,2^{x+3}-2^{x+2}-2^{x+1}=64\)
\(\Rightarrow2^x.8-2^x.4-2^x.2=64\)
\(\Rightarrow2^x.2=64\Rightarrow2^x=32\Rightarrow x=5\)
Vậy x = 5
b, \(\left(x-1\right)^4=\left(x-1\right)^2\)
\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
+) \(\left(x-1\right)^2=0\Rightarrow x=1\)
+) \(\left(x-1\right)^2-1=0\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy...