$$\eqalign{
& a)\,\,{{{x^2}} \over 3} = {9 \over x}\,\left( {x \ne 0} \right) \cr
& {x^3} = 3.9 \cr
& {x^3} = 27 \cr
& {x^3} = {3^3} \cr
& x = 3 \cr
& b)\,\,\left| {2.{1 \over 2} - x} \right| = 4 \cr
& \,\,\,\,\,\,\left| {1 - x} \right| = 4 \cr
& \,\,\,\,\,\,1 - x = 4\,\,hoac\,\,1 - x = - 4 \cr
& \,\,\,\,\,x = - 3\,\,hoac\,x = 5 \cr
& c)\,\,{3 \over 4} - \left( {x + {1 \over 2}} \right) = 1{2 \over 3} \cr
& \,\,\,\,\,{3 \over 4} - \left( {x + {1 \over 2}} \right) = {5 \over 3} \cr
& \,\,\,\,\,x + {1 \over 2} = {3 \over 4} - {5 \over 3} \cr
& \,\,\,\,\,x + {1 \over 2} = - {{11} \over {12}} \cr
& \,\,\,\,\,x = - {{11} \over {12}} - {1 \over 2} \cr
& \,\,\,\,\,x = - {{17} \over {12}} \cr
& d)\,\,{\left( {x - {1 \over 3}} \right)^3} = {1 \over {125}} \cr
& \,\,\,\,\,\,\,{\left( {x - {1 \over 3}} \right)^3} = {\left( {{1 \over 5}} \right)^3} \cr
& \,\,\,\,\,\,\,x - {1 \over 3} = {1 \over 5} \cr
& \,\,\,\,\,\,\,\,x = {1 \over 5} + {1 \over 3} \cr
& \,\,\,\,\,\,\,\,x = {8 \over {15}} \cr} $$
