Cho số nguyên dương \(n \ge 2\) và các số thực \(a,b\), nếu có \({a^n} = b\) thì:

\(a > 0\)

\(a = 0\)

\(a \ne 0\)

\(a < 0\)

\({a^n}\)         

\({n^a}\)

\(na\)

\(a + n\)

\({a^{\dfrac{3}{2}}} = \sqrt {{a^3}} \)          

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

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

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

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

\({a^{\dfrac{1}{{10}}}} = \sqrt {{a^{10}}} \)

\({a^{\dfrac{1}{{10}}}} = {a^{10}}\)

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

\({a^{545}} = {a^5}.{a^9}\)

\({a^{45}} = {a^5} + {a^9}\)

\({a^{45}} = {a^9}:{a^5}\)

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

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

\({a^3}.{a^2} = {a^6}\)

\({a^6} = {a^3} + {a^3}\)

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

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

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

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

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