1.
\(\left(x+\dfrac{1}{3}\right)^2=16\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=4\\x+\dfrac{1}{3}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{-13}{3}\end{matrix}\right.\)
2.
\(x+\dfrac{3}{10}=\dfrac{7}{5}\\ x=\dfrac{7}{5}-\dfrac{3}{10}\\ x=\dfrac{11}{10}\)
\(\left(x+\dfrac{1}{3}\right)^2=16\)
\(\left[{}\begin{matrix}\left(x+\dfrac{1}{3}\right)^2=4^2\\\left(x+\dfrac{1}{3}\right)^2=\left(-4\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+\dfrac{1}{3}=4\\x+\dfrac{1}{3}=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4-\dfrac{1}{3}\\x=-4-\dfrac{1}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{-13}{3}\end{matrix}\right.\)
\(x+\dfrac{3}{10}=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}-\dfrac{3}{10}\)
\(x=\dfrac{14}{10}-\dfrac{3}{10}\)
\(x=\dfrac{11}{10}\)
Vậy x = \(\dfrac{11}{10}\)