Số $x$ sao cho ${2^x}\; = {\left( {{2^2}} \right)^5}$   là :

${a^m}.{a^n} = {a^{m + n}}$

${\left( {a.b} \right)^m} = {a^m}.{b^m}$

\({\left( {{a^m}} \right)^n} = {a^{m + n}}\)

${\left( {{a^m}} \right)^n} = {a^{m.n}}$

$\dfrac{8}{9}$

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

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

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

${x^0} = x$

${x^1} = 1$    

\({x^0} = 1\)

${\left( {\dfrac{x}{y}} \right)^n} = \dfrac{{{x^n}}}{y^n}\left( {y \ne 0;\,n \in \mathbb{N}} \right)$

$7$

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

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

$1$

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

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

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

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

${x^{18}}:{x^6}(x\; \ne 0)$

${x^4}.{\rm{ }}{x^8}$

${x^2}.{\rm{ }}{x^6}$

${\left( {{x^3}} \right)^4}$

${8^8}$

${9^8}$          

${6^8}$       

Một đáp số khác