Lũy thừa với số mũ hữu tỉ

Ta có tính chất ${{\left( \dfrac{a}{b} \right)}^{x}}=\dfrac{{{a}^{x}}}{{{b}^{x}}}={{a}^{x}}{{b}^{-x}}.$

Ta có: $A=\dfrac{{{a}^{\dfrac{4}{3}}}-8{{a}^{\dfrac{1}{3}}}b}{{{a}^{\dfrac{2}{3}}}+2\sqrt[3]{ab}+4{{b}^{\dfrac{2}{3}}}}.{{\left( 1-2\sqrt[3]{\dfrac{b}{a}} \right)}^{-1}}-{{a}^{\dfrac{2}{3}}}$

$=\dfrac{{{a}^{\dfrac{1}{3}}}\left( a-8b \right)}{{{a}^{\dfrac{2}{3}}}+2{{\left( ab \right)}^{\dfrac{1}{3}}}+4{{b}^{\dfrac{2}{3}}}}.{{\left( 1-2{{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right)}^{-1}}-{{a}^{\dfrac{2}{3}}}$

Đặt $\left\{ \begin{matrix} x={{a}^{\dfrac{1}{3}}} \\ y={{b}^{\dfrac{1}{3}}} \\ \end{matrix}\Rightarrow A=\dfrac{x\left( {{x}^{3}}-8{{y}^{3}} \right)}{{{x}^{2}}+2xy+4{{y}^{2}}}{{\left( \dfrac{x-2y}{x} \right)}^{-1}}-{{x}^{2}} \right.$ . $=\dfrac{x\left( {{x}^{3}}-8{{y}^{3}} \right)}{{{x}^{2}}+2xy+4{{y}^{2}}}.\dfrac{x}{x-2y}-{{x}^{2}}=\dfrac{{{x}^{2}}\left( {{x}^{3}}-8{{y}^{3}} \right)}{{{x}^{3}}-8{{y}^{3}}}-{{x}^{2}}=0$

Ta có: $\dfrac{{{x}^{\dfrac{5}{4}}}y+x{{y}^{\dfrac{5}{4}}}}{\sqrt[4]{x}+\sqrt[4]{y}}=\dfrac{xy\left( {{x}^{\dfrac{1}{4}}}+{{y}^{\dfrac{1}{4}}} \right)}{\sqrt[4]{x}+\sqrt[4]{y}}=xy$

${K = {{\left( {{x^{\frac{1}{2}}} - {y^{\frac{1}{2}}}} \right)}^2}{{\left( {1 - 2\sqrt {\frac{y}{x}}  + \frac{y}{x}} \right)}^{ - 1}}}$

${ = {{\left( {\sqrt x  - \sqrt y } \right)}^2}{{\left( {\frac{{x - 2\sqrt {xy}  + y}}{x}} \right)}^{ - 1}}}$

${ = {{\left( {\sqrt x  - \sqrt y } \right)}^2}\frac{x}{{{{\left( {\sqrt x  - \sqrt y } \right)}^2}}} = x}$