Tính $A = \,\sqrt[3]{{2 + 10\sqrt {\dfrac{1}{{27}}} }}\, + \,\sqrt[3]{{2 - 10\sqrt {\dfrac{1}{{27}}} }}$

Ta có:

$\begin{array}{l}A = \,\sqrt[3]{{2 + 10\sqrt {\dfrac{1}{{27}}} }}\, + \,\sqrt[3]{{2 - 10\sqrt {\dfrac{1}{{27}}} }}\\{A^3} = {\left( {\sqrt[3]{{2 + 10\sqrt {\dfrac{1}{{27}}} }}\, + \,\sqrt[3]{{2 - 10\sqrt {\dfrac{1}{{27}}} }}} \right)^3}\\\,\,\,\,\,\,\, = \,2 + 10\sqrt {\dfrac{1}{{27}}}  + 2 - 10\sqrt {\dfrac{1}{{27}}}  + 3.\sqrt[3]{{2 + 10\sqrt {\dfrac{1}{{27}}} }}.\sqrt[3]{{2 - 10\sqrt {\dfrac{1}{{27}}} }}.\left( {\sqrt[3]{{2 + 10\sqrt {\dfrac{1}{{27}}} }}\, + \,\sqrt[3]{{2 - 10\sqrt {\dfrac{1}{{27}}} }}} \right)\\\,\,\,\,\,\,\, = \,4 + 3.\sqrt[3]{{{2^2} - {{\left( {10\sqrt {\dfrac{1}{{27}}} } \right)}^2}}}.A\\\,\,\,\,\,\,\, = \,4 + 3.\sqrt[3]{{\dfrac{8}{{27}}}}.A\\\,\,\,\,\,\,\, = \,4 + 3.\dfrac{2}{3}.A\\\,\,\,\,\,\,\, = 4 + 2A\end{array}$

Vậy giá trị của A thảo mãn phương trình ${A^3} = 4 + 2A$

$\begin{array}{l} \Leftrightarrow {A^3} - 2A - 4 = 0\\ \Leftrightarrow {A^3} - 8 - 2A + 4 = 0\\ \Leftrightarrow \left( {A - 2} \right)\left( {{A^2} + 2A + 4} \right) - 2\left( {A - 2} \right) = 0\\ \Leftrightarrow \left( {A - 2} \right)\left( {{A^2} + 2A + 4 - 2} \right) = 0\\ \Leftrightarrow \left( {A - 2} \right)\left( {{A^2} + 2A + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}A - 2 = 0\\{A^2} + 2A + 2 = 0\,\,\left( {vô\,\,nghiệm} \right)\end{array} \right. \Leftrightarrow A = 2.\end{array}$

(Do ${A^2} + 2A + 2 = {\left( {A + 1} \right)^2} + 1 > 0$ với mọi A).

Vậy giá trị của $A = 2$.