Задача 59. Вычислите \(\dfrac{{4-3\sqrt 2 }}{{{{\left( {\sqrt[4]{2}-\sqrt[4]{8}} \right)}^2}}} + \dfrac{{{{\left( {\sqrt[3]{9} + \sqrt 3 } \right)}^2}}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}}\)
ОТВЕТ: 2.
\(\dfrac{{4-3\sqrt 2 }}{{{{\left( {\sqrt[4]{2}-\sqrt[4]{8}} \right)}^2}}} + \dfrac{{{{\left( {\sqrt[3]{9} + \sqrt 3 } \right)}^2}}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}}.\) Воспользуемся формулами сокращённого умножения: \({\left( {a-b} \right)^2} = {a^2}-2ab + {b^2};\) \({\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}.\) \(\dfrac{{4-3\sqrt 2 }}{{{{\left( {\sqrt[4]{2}-\sqrt[4]{8}} \right)}^2}}} + \dfrac{{{{\left( {\sqrt[3]{9} + \sqrt 3 } \right)}^2}}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}} = \dfrac{{4-3\sqrt 2 }}{{{{\left( {\sqrt[4]{2}} \right)}^2}-2\sqrt[4]{2} \cdot \sqrt[4]{8} + {{\left( {\sqrt[4]{8}} \right)}^2}}} + \dfrac{{{{\left( {\sqrt[3]{9}} \right)}^2} + 2\sqrt[3]{9} \cdot \sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}} = \) \( = \dfrac{{4-3\sqrt 2 }}{{\sqrt 2 -2\sqrt[4]{{16}} + \sqrt 8 }} + \dfrac{{\sqrt[3]{{{3^3} \cdot 3}} + 2\sqrt[6]{{{3^4}}} \cdot \sqrt[6]{{{3^3}}} + 3}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}} = \)\( \dfrac{{4-3\sqrt 2 }}{{\sqrt 2 -2 \cdot 2 + 2\sqrt 2 }} + \dfrac{{3\sqrt[3]{3} + 2\sqrt[6]{{{3^7}}} + 3}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}} = \) \( = \dfrac{{4-3\sqrt 2 }}{{-\left( {4-3\sqrt 2 } \right)}} + \dfrac{{3\sqrt[3]{3} + 2 \cdot 3 \cdot \sqrt[6]{3} + 3}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}} = -1 + \dfrac{{3\left( {\sqrt[3]{3} + 2\sqrt[6]{3} + 1} \right)}}{{\sqrt[3]{3} + 2\sqrt[6]{3} + 1}} = -1 + 3 = 2.\) Ответ: 2.