Cho ${20^n}\;:\;{5^n} = 4$ thì  :

\(x = 0\)

$x =  \pm \;\dfrac{1}{2}$

$x = \;\dfrac{1}{2}$

\(x =  - \dfrac{1}{2}\)

$\dfrac{8}{9}$

$\dfrac{8}{{27}}$

\(\dfrac{4}{9}\)

$\dfrac{4}{{27}}$

$1$ số

$2$ số

$0$ số

$3$ số

$7$

$\dfrac{1}{{49}}$

\(\dfrac{1}{7}\)

$1$

$\left| {-{\rm{ }}0,4} \right|{\rm{ }} = {\rm{ }}0,4$

$\left| {-{\rm{ }}0,4} \right|{\rm{ }} = -{\rm{ }}0,4$

$\left| {-\,0,4} \right| =  \pm {\rm{ }}0,4$

$\left| {-{\rm{ }}0,4} \right| = 0$

${\left( {-2019} \right)^0} = 1$

$\left( {0,5} \right).{\left( {0,5} \right)^2} = \dfrac{1}{4}$

${4^6}:{\rm{ }}{4^4} = 16$

${\left( {-3} \right)^3}.{\left( {-{\rm{ }}3} \right)^{{\rm{ }}2}} = {\left( { - 3} \right)^5}$

\(1,7\)

\( - 0,2\)

\(0,2\)

\(2,8\)