Step-by-step explanation:
For each case we have the next step by step solution.
- [tex]x^2(\dfrac{x-2}{3})=\dfrac{x^3-2x^2}{3}[/tex]
- [tex]\dfrac{x^{1/4}}{x-\dfrac{5}{2}}=\dfrac{x^{1/4}}{\dfrac{2x}{2}-\dfrac{5}{2}}=\dfrac{x^{1/4}}{\dfrac{2x-5}{2}}=\dfrac{2x^{1/4}}{{2x-5}}[/tex]
- [tex]\dfrac{\dfrac{4}{5}x-\dfrac{2}{5}y^{3/2}}{\dfrac{\dfrac{2}{3}x^3}{5y^{1/2}}}=\dfrac{\dfrac{4x}{5}-\dfrac{2y^{3/2}}{5}}{\dfrac{\dfrac{2x^3}{3}}{5y^{1/2}}}=\dfrac{\dfrac{4x-2y^{3/2}}{5}}{\dfrac{2x^3}{15y^{1/2}}}={\dfrac{(4x-2y^{3/2})\cdot 15y^{1/2}}{5\cdot 2x^3}}[/tex] [tex]{\dfrac{(4x-2y^{3/2})\cdot 15y^{1/2}}{5\cdot 2x^3}}={\dfrac{(60xy^{1/2}-30y^{3/2}y^{1/2})}{10x^3}}={\dfrac{(60xy^{1/2}-30y^{4/2})}{10x^3}}={\dfrac{(60xy^{1/2}-30y^{2})}{10x^3}}[/tex]
- [tex]\dfrac{\dfrac{\dfrac{2}{3}x^2}{3y^{2/3}}}{\dfrac{1}{3}x-\dfrac{1}{3}y-\dfrac{1}{3}}=\dfrac{\dfrac{\dfrac{2x^2}{3}}{3y^{2/3}}}{\dfrac{x}{3}-\dfrac{y}{3}-\dfrac{1}{3}}=\dfrac{\dfrac{2x^2}{9y^{2/3}}}{\dfrac{x-y-1}{3}}=\dfrac{2x^2\cdot 3}{(x-y-1)\cdot 9y^{2/3}}}=\dfrac{6x^2}{(9xy^{2/3}-9yy^{2/3}-9y^{2/3})}}=\dfrac{6x^2}{(9xy^{2/3}-9y^{5/3}-9y^{2/3})}}[/tex]
- [tex](z^{2/3}x^{2/3}y+\dfrac{2}{3})y+(z^{2/3}x-\dfrac{1}{3}y-\dfrac{1}{3})x=(z^{2/3}x^{2/3}y^2+\dfrac{2}{3}y)+(z^{2/3}x^2-\dfrac{1}{3}yx-\dfrac{x}{3})=z^{2/3}x^{2/3}y^2+z^{2/3}x^2-\dfrac{1}{3}yx+\dfrac{2}{3}y-\dfrac{x}{3}[/tex]
- [tex](\dfrac{4x^3}{5y^2}z^2)^{1/3}=\dfrac{(4x^3)^{1/3}}{(5y^2)^{1/3}}(z^2)^{1/3}=\dfrac{4^{1/3}x}{5^{1/3}y^{2/3}}z^{2/3}[/tex]
