Cho $n \in Z,n < 0$, đẳng thức ${a^n} = \dfrac{1}{{{a^{ - n}}}}$ xảy ra khi:

${a^n}$         

${n^a}$

$na$

$a + n$

${a^{\dfrac{3}{2}}} = \sqrt {{a^3}}$          

${a^{\dfrac{3}{2}}} = \sqrt[3]{{{a^2}}}$

${a^{\dfrac{3}{2}}} = \sqrt[6]{a}$

${a^{\dfrac{3}{2}}} = \sqrt[3]{{{a^6}}}$

${a^{\dfrac{1}{{10}}}} = \sqrt[{10}]{a}$

${a^{\dfrac{1}{{10}}}} = \sqrt {{a^{10}}}$

${a^{\dfrac{1}{{10}}}} = {a^{10}}$

${a^{\dfrac{1}{{10}}}} = \sqrt[a]{{10}}$

${a^{545}} = {a^5}.{a^9}$

${a^{45}} = {a^5} + {a^9}$

${a^{45}} = {a^9}:{a^5}$

${a^{45}} = {\left( {{a^9}} \right)^5}$

${\left( {{a^3}} \right)^2} = {a^3}.{a^2}$

${a^3}.{a^2} = {a^6}$

${a^6} = {a^3} + {a^3}$

${\left( {{a^3}} \right)^2} = {\left( {{a^2}} \right)^3}$

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

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

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

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