Cho \(n \in {\mathbb{N}^*}\), với điều kiện nào dưới đây của \(x\) thì đẳng thức \(\sqrt[n]{x} = {x^{\dfrac{1}{n}}}\) luôn xảy ra?

\(D = \mathbb{R}\backslash \left\{ { - 1;4} \right\}\).     

\(D = \mathbb{R}\).

\(D = \left( { - \infty ; - 1} \right) \cup \left( {4; + \infty } \right)\).

\(D = \left( { - \infty ; - 1} \right] \cup \left[ {4; + \infty } \right)\).

\(y = {x^5}\)

\(y = {x^{ - 1}}\)

\(y = {x^{\sqrt 2 }}\)

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

\(y' = \dfrac{4}{3}{x^{ - \frac{1}{3}}}\)

\(y' = \dfrac{4}{3}{x^{\frac{1}{3}}}\)

\(y' = \dfrac{3}{7}{x^{\frac{7}{3}}}\)

\(y' = \dfrac{3}{4}{x^{\frac{1}{3}}}\)

\(\dfrac{5}{3}{\left( {2x} \right)^{\dfrac{2}{3}}}\)

\(\dfrac{{10}}{3}{\left( {2x} \right)^{\dfrac{2}{3}}}\)

\(\dfrac{5}{6}{\left( {2x} \right)^{\dfrac{2}{3}}}\)

\(\dfrac{{10}}{3}{\left( {2x} \right)^{\dfrac{5}{3}}}\)

\(\left( {\sqrt[n]{x}} \right)' = \dfrac{1}{n}{x^{ - \dfrac{{n - 1}}{n}}}\)

\(\left( {\sqrt[n]{x}} \right)' = \dfrac{1}{n}{x^{\dfrac{{n - 1}}{n}}}\)         

\(\left( {\sqrt[n]{x}} \right)' = n{x^{ - \dfrac{{n - 1}}{n}}}\)            

\(\left( {\sqrt[n]{x}} \right)' = \dfrac{1}{n}{x^{ - \dfrac{{n + 1}}{n}}}\)

\(y = {x^{ - 4}}\).

\(y = {x^4}\).

$y = {x^{ - \dfrac{3}{4}}}$.

$y = \sqrt[3]{x}$.