Tìm $x$ để biểu thức \({\left( {{x^2} - 1} \right)^{\sqrt {\dfrac{1}{3}} }}\) có nghĩa:

\(x <  - 1\)

\(x > 1\)

\(x \in R\)

\(x \ge 1\)

\(\forall x \in \mathbb{R}\)

Không tồn tại \(x\)

\(\forall x > 1\)

\(\forall x \in \mathbb{R}\backslash \left\{ {\rm{0}} \right\}\)

$a >  - 2$

$\forall a \in \mathbb{R}$

$a > 0$

$a <  - 2$

\({\left( {1 - \sqrt 2 } \right)^{ - 3}}\)

\({\left( {2 - \sqrt 2 } \right)^0}\)

\({\left( {3 - 2\sqrt 2 } \right)^{3 - 2\sqrt 2 }}\)

\({\left( { - \dfrac{3}{2}} \right)^\pi }\)

${\left( { - 3} \right)^{ - 4}}$.

${\left( { - 3} \right)^{ - \dfrac{1}{3}}}$.

${0^4}$.

${\left( {\dfrac{1}{{{2^{ - 3}}}}} \right)^0}$.

${a^{\dfrac{1}{n}}} = \sqrt[n]{a}$,$\forall a \ne 0$.

${a^{\dfrac{1}{n}}} = \sqrt[n]{a}$,$\forall a > 0$.

${a^{\dfrac{1}{n}}} = \sqrt[n]{a}$,$\forall a \ge 0$.

${a^{\dfrac{1}{n}}} = \sqrt[n]{a}$,$\forall a \in \mathbb{R}$.