Giải câu 2 bài: Lũy thừa.
Áp dụng công thức lũy thừa, ta có:
a) $a^{\frac{1}{3}}.\sqrt{a}$
= $a^{\frac{1}{3}}.a^{\frac{1}{2}}$
= $a^{\frac{1}{3}+\frac{1}{2}}=a^{\frac{5}{6}}$
Vậy $a^{\frac{1}{3}}.\sqrt{a}=a^{\frac{5}{6}}$
b) $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}$
= $b^{\frac{1}{2}}.b^{\frac{1}{3}}.b^{\frac{1}{6}}$
= $b^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=b$
Vậy $b^{\frac{1}{2}}.b^{\frac{1}{3}}.\sqrt[6]{b}=b$
c) $a^{\frac{4}{3}}:\sqrt[3]{a}$
= $a^{\frac{4}{3}}:a^{\frac{1}{3}}$
= $a^{\frac{4}{3}-\frac{1}{3}}=a$
Vậy $a^{\frac{4}{3}}:\sqrt[3]{a}=a$
d) $\sqrt[3]{b}:b^{\frac{1}{6}}$
= $b^{\frac{1}{3}}:b^{\frac{1}{6}}$
= $b^{\frac{1}{3}-\frac{1}{6}}=b^{\frac{1}{6}}$
Vậy $\sqrt[3]{b}:b^{\frac{1}{6}}=b^{\frac{1}{6}}$