Bài 1. Ta có: \(\root 3 \of {{a \over {{b^2}}}} = \root 3 \of {{{ab} \over {{b^3}}}} = {{\root 3 \of {ab} } \over {\root 3 \of {{b^3}} }} = {1 \over b}\root 3 \of {ab} \,\,\left( {đpcm} \right)\)
Bài 2. Ta có:
\(\eqalign{ & \root 3 \of {x - 5} + 3 = 0 \Leftrightarrow \root 3 \of {x - 5} = - 3 \cr & \Leftrightarrow {\left( {\root 3 \of {x - 5} } \right)^3} = {\left( { - 3} \right)^3}\cr& \Leftrightarrow x - 5 = - 27 \cr & \Leftrightarrow x = - 22 \cr} \)
Bài 3. Ta có:
\(2\root 3 \of 3 > \root 3 \of {23} \Leftrightarrow {\left( {2\root 3 \of 3 } \right)^3} > {\left( {\root 3 \of {23} } \right)^3} \)\(\Leftrightarrow {2^3}.3 > 23\)
\(⇔ 24 > 23\) (luôn đúng)
Bài 4. Ta có: \({1 \over {2\root 3 \of 2 }} = {{1{{\left( {\root 3 \of 2 } \right)}^2}} \over {2{{\left( {\root 3 \of 2 } \right)}^3}}} = {{\root 3 \of 4 } \over 4}\)
