\(a,\left(4x-\dfrac{1}{3}\right)^2+\dfrac{26}{25}=3\\ \Leftrightarrow\left(4x-\dfrac{1}{3}\right)^2=\dfrac{49}{25}\\ \Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{1}{3}=\dfrac{7}{5}\\4x-\dfrac{1}{3}=-\dfrac{7}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{30}\\x=-\dfrac{4}{15}\end{matrix}\right.\)
\(b,\left(2-\dfrac{x}{5}\right)^3-\dfrac{20}{9}=\dfrac{4}{27}\\ \Leftrightarrow\left(2-\dfrac{x}{5}\right)^3=\dfrac{64}{27}\\ \Leftrightarrow2-\dfrac{x}{5}=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{10}{3}\)
\(c,\left|3x+\dfrac{1}{3}\right|-\dfrac{1}{2}=4\\\Leftrightarrow\left|3x+\dfrac{1}{3}\right|=\dfrac{9}{2}\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=\dfrac{9}{2}\left(3x+\dfrac{1}{3}\ge0\right)\\3x+\dfrac{1}{3}=-\dfrac{9}{2}\left(3x+\dfrac{1}{3}< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{25}{18}\left(x\ge-\dfrac{1}{9}\right)\left(N\right)\\x=-\dfrac{29}{18}\left(x< -\dfrac{1}{9}\right)\left(N\right)\end{matrix}\right. \)
