Cho số thực \[a\] dương. Khi đó

a) \({a^{\frac{3}{2}}}.{a^{\frac{2}{9}}} = {a^{\frac{3}{2} + \frac{2}{9}}} = {a^{\frac{{31}}{{18}}}}\). Vậy \({a^{\frac{3}{2}}}.{a^{\frac{2}{9}}} = \sqrt[3]{a}\) (sai).

b) \({a^{\frac{3}{2}}}.\sqrt a  = {a^{\frac{3}{2}}}.{a^{\frac{1}{2}}} = {a^{\frac{3}{2} + \frac{1}{2}}} = {a^2}\). Vậy \({a^{\frac{3}{2}}}.\sqrt a  = {a^2}\) (đúng).

c) \[\sqrt[3]{a}.\sqrt[6]{a} = {a^{\frac{1}{3}}}.{a^{\frac{1}{6}}} = {a^{\frac{1}{2}}} = \sqrt a \]. Vậy \[\sqrt[3]{a}.\sqrt[6]{a} = \sqrt a \] (đúng).

d) Ta có: \[{a^{\sqrt 5 }}{\left( {\frac{1}{a}} \right)^{\sqrt 5  - 2}} = {a^{\sqrt 5 }}.{a^{ - \left( {\sqrt 5  - 2} \right)}} = {a^{\sqrt 5 }}.{a^{2 - \sqrt 5 }} = {a^{\sqrt 5  + 2 - \sqrt 5 }} = {a^2}.\] Vậy \[{a^{\sqrt 5 }}.{\left( {\frac{1}{a}} \right)^{\sqrt 5  - 2}} = {a^{ - 2}}\] (sai).