\(a,\\ \left(x+\dfrac{1}{2}\right)^3=\dfrac{8}{125}=\dfrac{2^3}{5^3}\\ \left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow\left(x+\dfrac{1}{2}\right)=\dfrac{2}{5}\\ x=\dfrac{2}{5}-\dfrac{1}{2}\\ x=-\dfrac{1}{10}\)
\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)
\(c,\\ \dfrac{1}{2}.\left(x+\dfrac{1}{3}\right)=\dfrac{1}{16}\\ \left(x+\dfrac{1}{3}\right)=\dfrac{1}{16}:\dfrac{1}{2}\\ \left(x+\dfrac{1}{3}\right)=\dfrac{1}{8}\\ x=\dfrac{1}{8}-\dfrac{1}{3}\\ x=-\dfrac{5}{24}\)
