Bài 2 trang 33 SGK Toán 11 Tập 2 Cánh diều - CD

a) Ta có: $a^{\frac{1}{3}}\cdot \sqrt{a}=a^{\frac{1}{3}}.a^{\frac{1}{2}}=a^{\frac{1}{3}+\frac{1}{2}}=a^{\frac{5}{6}}=\sqrt[6]{a^5};$

b) Tâ có: $b^{\frac{1}{2}}\cdot b^{\frac{1}{3}}\cdot \sqrt[6]{b}=b^{\frac{1}{2}}\cdot b^{\frac{1}{3}}\cdot b^{\frac{1}{6}}=b^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=b^1;$

c) Ta có: $a^{\frac{4}{3}}:\sqrt[3]{a}=a^{\frac{4}{3}}:a^{\frac{1}{3}}=a^{\frac{4}{3}-\frac{1}{3}}=a^1;$

d) Ta có: $\sqrt[3]{b}:b^{\frac{1}{6}}=b^{\frac{1}{3}}:b^{\frac{1}{6}}=b^{\frac{1}{3}-\frac{1}{6}}=b^{\frac{1}{6}}=\sqrt[6]{b}.$

-- Mod Toán 11 HỌC247