\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\frac{-14}{5}\right|\)
\(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\-\frac{5}{3}\end{cases}}}\)
Vậy...
\(\Rightarrow\left(x-7\right)^{x+1}.\left(1-\left(x-7\right)^{10}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1=1^{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=6;x=8\end{cases}}\)