trục căn thức 1/∛16+∛12+∛9

Đáp án:

\(\root 3 \of 4 - \root 3 \of 3 \)

Giải thích các bước giải:

$$\eqalign{ & A = {1 \over {\root 3 \of {16} + \root 3 \of {12} + \root 3 \of 9 }} \cr & = {1 \over {{{\left( {\root 3 \of 4 } \right)}^2} + \root 3 \of 4 .\root 3 \of 3 + {{\left( {\root 3 \of 3 } \right)}^2}}} \cr & = {{\root 3 \of 4 - \root 3 \of 3 } \over {\left( {\root 3 \of 4 - \root 3 \of 3 } \right)\left[ {{{\left( {\root 3 \of 4 } \right)}^2} + \root 3 \of 4 .\root 3 \of 3 + {{\left( {\root 3 \of 3 } \right)}^2}} \right]}} \cr & = {{\root 3 \of 4 - \root 3 \of 3 } \over {{{\left( {\root 3 \of 4 } \right)}^3} - {{\left( {\root 3 \of 3 } \right)}^3}}} = {{\root 3 \of 4 - \root 3 \of 3 } \over {4 - 3}} = \root 3 \of 4 - \root 3 \of 3 \cr} $$