Cho $a > 0,b < 0$, khẳng định nào sau đây là khẳng định sai?

$x <  - 1$

$x > 1$

$x \in R$

$x \ge 1$

$\forall x \in \left( { - \infty ;-1} \right] \cup \left[ {1; + \infty } \right)$.

$\forall x \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)$.

$\forall x \in \left( { - 1;1} \right)$.

$\forall x \in \mathbb{R}\backslash \left\{ { \pm 1} \right\}$.

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

Không tồn tại $x$

$\forall x > 1$

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

$a >  - 2$

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

$a > 0$

$a <  - 2$

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

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

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

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

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

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

${0^4}$.

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

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

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

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

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