Bài giải

a, \(\left(3x-1\right)\left(x+1\right)>0\)

Khi  \(\orbr{\begin{cases}3x-1< 0\\x+1< 0\end{cases}}\Rightarrow\orbr{\begin{cases}3x< 1\\x< -1\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}\\x< -1\end{cases}}\)

Hoặc \(\orbr{\begin{cases}3x-1>0\\x+1>0\end{cases}}\Rightarrow\orbr{\begin{cases}3x>1\\x>-1\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{1}{3}\\x>-1\end{cases}}\)

b, \(\left(x+2\right)^2\left(x-3\right)\le0\)

\(\Rightarrow\text{ }\left(x+2\right)^2\text{ và }\left(x-3\right)\) đối nhau

Mà \(\left(x+2\right)^2\ge0\) nên \(\hept{\begin{cases}\left(x+2\right)^2\ge0\\x-3\le0\end{cases}}\Rightarrow\hept{\begin{cases}x+2\ge0\\x\le3\end{cases}}\Rightarrow\hept{\begin{cases}x\ge-2\\x\le3\end{cases}}\text{ }\left(\text{ loại}\right)\)

\(\Rightarrow\text{ }x\in\varnothing\)

c, \(\left(x-\frac{1}{3}\right)^5=4\left(x-\frac{1}{3}\right)^3\)

\(\left(x-\frac{1}{3}\right)^5-4\left(x-\frac{1}{3}\right)^3=0\)

\(\left(x-\frac{1}{3}\right)^3\left[\left(x-\frac{1}{3}\right)^2-4\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^3=0\\\left(x-\frac{1}{3}\right)^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=0\\\left(x-\frac{1}{3}\right)^2=4=\left(\pm2\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{3}\text{ ; }x=\frac{7}{3}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{\frac{1}{3}\text{ ; }-\frac{5}{3}\text{ ; }\frac{7}{3}\right\}\)