Định nghĩa.
Chọn đáp án $ {{\left( {{10}^{\alpha }} \right)}^{2}}={{10}^{{{\alpha }^{2}}}}. $
Ta có $ \left( \sqrt{x}-\sqrt[4]{x}+1 \right)\left( \sqrt{x}+\sqrt[4]{x}+1 \right)\left( x-\sqrt{x}+1 \right)=\left[ {{\left( \sqrt{x}+1 \right)}^{2}}-\sqrt{x} \right]\left( x-\sqrt{x}+1 \right) $
$ =\left( x+\sqrt{x}+1 \right)\left( x-\sqrt{x}+1 \right)={{\left( x+1 \right)}^{2}}-x={{x}^{2}}+x+1 $ .
Ta có $\sqrt[4]{{{\left( ab \right)}^{4}}}=\left| ab \right|=-ab\left( a<0,b>0 \right)$ nên khẳng định “$\sqrt[4]{{{\left( ab \right)}^{4}}}=ab$” sai
Theo tính chất của lũy thừa.
Ta có: $ \dfrac{{{a}^{\frac{1}{3}}}{{b}^{\frac{-1}{3}}}-{{a}^{-\frac{1}{3}}}{{b}^{\frac{1}{3}}}}{\sqrt[3]{{{a}^{2}}}-\sqrt[3]{{{b}^{2}}}}=\dfrac{\dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}-\dfrac{\sqrt[3]{b}}{\sqrt[3]{a}}}{\sqrt[3]{{{a}^{2}}}-\sqrt[3]{{{b}^{2}}}}=\dfrac{\dfrac{\sqrt[3]{{{a}^{2}}}-\sqrt[3]{{{b}^{2}}}}{\sqrt[3]{ab}}}{\sqrt[3]{{{a}^{2}}}-\sqrt[3]{{{b}^{2}}}}=\dfrac{1}{\sqrt[3]{ab}} $ .
Ta có: $ \dfrac{{{a}^{\sqrt{7}+1}}.{{a}^{2-\sqrt{7}}}}{{{\left( {{a}^{\sqrt{2}-2}} \right)}^{\sqrt{2}+2}}}=\dfrac{{{a}^{3}}}{{{a}^{-2}}}={{a}^{5}} $ .