Bài 1. Biến đổi vế trái (VT), ta có:
\(\eqalign{ VT &= \left[ {{{\left( {\root 3 \of 3 } \right)}^2} + \root 3 \of 3 .\root 3 \of 2 + {{\left( {\root 3 \of 2 } \right)}^2}} \right].\left[ {\root 3 \of 3 - \root 3 \of 2 } \right] \cr & = {\left( {\root 3 \of 3 } \right)^3} - {\left( {\root 3 \of 2 } \right)^3} \cr&= 3 - 2 = 1 = VP\,\left( {đpcm} \right) \cr} \)
Bài 2. Ta có:
\(\eqalign{ & \root 3 \of {{x^3} + 8} = x + 2 \cr&\Leftrightarrow {x^3} + 8 = {\left( {x + 2} \right)^3} \cr & \Leftrightarrow {x^3} + 8 = {x^3} + 6{x^2} + 12x + 8 \cr & \Leftrightarrow 6{x^2} + 12x = 0 \Leftrightarrow x\left( {x + 2} \right) = 0 \cr & \Leftrightarrow \left[ {\matrix{ {x = 0} \cr {x = - 2} \cr } } \right. \cr} \)
Bài 3. Ta có:
\(\eqalign{ & 3\root 3 \of 3 > \root 3 \of {80} \Leftrightarrow {\left( {3\root 3 \of 3 } \right)^3} > {\left( {\root 3 \of {80} } \right)^3} \cr & \Leftrightarrow {3^3}{\left( {\root 3 \of 3 } \right)^3} > 80 \Leftrightarrow 27.3 > 80 \cr} \)
\(\;\;⇔ 81 > 80 \) (luôn đúng)
Bài 4. Ta có:
\(\eqalign{ & {1 \over {1 - \root 3 \of 2 }} \cr&= {{1 + \root 3 \of 2 + \root 3 \of 4 } \over {\left( {1 - \root 3 \of 2 } \right)\left( {1 + \root 3 \of 2 + \root 3 \of 4 } \right)}} \cr & = {{1 + \root 3 \of 2 + \root 3 \of 4 } \over {1 - 2}} = - \left( {1 + \root 3 \of 2 + \root 3 \of 4 } \right) \cr} \)
