Cho biểu thức \[P = \sqrt[3]{{x.\sqrt[4]{{{x^3}\sqrt x }}}}\], với \[x > 0.\] Mệnh đề nào dưới đây đúng?

\(A = \frac{{{{\left( {{{10}^2}} \right)}^\alpha } \cdot {{\left( {{{10}^{ - 3}}} \right)}^\beta }}}{{{{10}^{ - \alpha }} \cdot {{10}^{2\beta }}}} = \frac{{{{10}^{2\alpha }}}}{{{{10}^{ - \alpha }}}} \cdot \frac{{{{10}^{ - 3\beta }}}}{{{{10}^{2\beta }}}} = {10^{2\alpha  - ( - \alpha )}} \cdot {10^{ - 3\beta  - 2\beta }}\)

\( = {10^{3\alpha }} \cdot {10^{ - 5\beta }} = {\left( {{{10}^\alpha }} \right)^3} \cdot {\left( {{{10}^\beta }} \right)^{ - 5}} = {3^3} \cdot {7^{ - 5}} = \frac{{{3^3}}}{{{7^5}}} = \frac{{27}}{{16807}}{\rm{. }}\)

\(P = \frac{{\sqrt a  + \sqrt[4]{a}}}{{\sqrt[4]{a} + \sqrt[4]{b}}} - \frac{{\sqrt a  - \sqrt b }}{{\sqrt[4]{a} - \sqrt[4]{b}}} = \frac{{{{(\sqrt[4]{a})}^2} + \sqrt[4]{{ab}}}}{{\sqrt[4]{a} + \sqrt[4]{b}}} - \frac{{{{(\sqrt[4]{a})}^2} - {{(\sqrt[4]{b})}^2}}}{{\sqrt[4]{a} - \sqrt[4]{b}}}\)

\( = \frac{{\sqrt[4]{a}(\sqrt[4]{a} + \sqrt[4]{b})}}{{\sqrt[4]{a} + \sqrt[4]{b}}} - \frac{{(\sqrt[4]{a} - \sqrt[4]{b})(\sqrt[4]{a} + \sqrt[4]{b})}}{{\sqrt[4]{a} - \sqrt[4]{b}}} = \sqrt[4]{a} - (\sqrt[4]{a} + \sqrt[4]{b}) =  - \sqrt[4]{b}\)