\(\text{a},\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3\\ x+\dfrac{2}{3}=\dfrac{5}{4}\\ x=\dfrac{5}{4}-\dfrac{2}{3}=\dfrac{5.3-3.4}{12}=\dfrac{7}{12}\\ b,\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right)^3\\ x-\dfrac{1}{2}=\dfrac{2}{7}\\ x=\dfrac{2}{7}+\dfrac{1}{2}=\dfrac{2.2+1.7}{14}=\dfrac{11}{14}\)
\(a.\left(x+\dfrac{2}{3}\right)^3=\dfrac{125}{64}\\ < =>\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3\\ < =>x+\dfrac{2}{3}=\dfrac{5}{4}< =>x=\dfrac{5}{4}-\dfrac{2}{3}=\dfrac{7}{12}\)
\(b.\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{343}\\ < =>\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right)^3\\ < =>x-\dfrac{1}{2}=\dfrac{2}{7}< =>x=\dfrac{2}{7}+\dfrac{1}{2}=\dfrac{11}{14}\)
