\(\dfrac{{{{\left( {\sqrt[4]{{24}} + \sqrt[4]{6}} \right)}^2}}}{{4\sqrt 3 + 3\sqrt 6 }} + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{{{\left( {\sqrt 3 -\sqrt[4]{{45}}} \right)}^2}}}.\)
Воспользуемся формулами сокращённого умножения:
\({\left( {a-b} \right)^2} = {a^2}-2ab + {b^2};\) \({\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}.\)
\(\dfrac{{{{\left( {\sqrt[4]{{24}} + \sqrt[4]{6}} \right)}^2}}}{{4\sqrt 3 + 3\sqrt 6 }} + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{{{\left( {\sqrt 3 -\sqrt[4]{{45}}} \right)}^2}}} = \)
\( = \dfrac{{{{\left( {\sqrt[4]{{24}}} \right)}^2} + 2 \cdot \sqrt[4]{{24}} \cdot \sqrt[4]{6} + {{\left( {\sqrt[4]{6}} \right)}^2}}}{{4\sqrt 3 + 3\sqrt 6 }} + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{{{\left( {\sqrt 3 } \right)}^2}-2 \cdot \sqrt 3 \cdot \sqrt[4]{{45}} + {{\left( {\sqrt[4]{{45}}} \right)}^2}}} = \)
\( = \dfrac{{\sqrt {24} + 2 \cdot \sqrt[4]{{4 \cdot 6 \cdot 6}} + \sqrt 6 }}{{4\sqrt 3 + 3\sqrt 6 }} + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{3-2 \cdot \sqrt[4]{{9 \cdot 45}} + \sqrt {45} }} = \)
\( = \dfrac{{2\sqrt 6 + 2 \cdot \sqrt {2 \cdot 6} + \sqrt 6 }}{{4\sqrt 3 + 3\sqrt 6 }} + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{3-2 \cdot \sqrt[4]{{{3^4} \cdot 5}} + \sqrt {9 \cdot 5} }} = \)
\( = \dfrac{{3\sqrt 6 + 4\sqrt 3 }}{{4\sqrt 3 + 3\sqrt 6 }} + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{3-6\sqrt[4]{5} + 3\sqrt 5 }} = 1 + 3 \cdot \dfrac{{1-2\sqrt[4]{5} + \sqrt 5 }}{{3\left( {1-2\sqrt[4]{5} + \sqrt 5 } \right)}} = 1 + 1 = 2.\)
Ответ: 2.