Задача 90. Вычислите \(\dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt[4]{{11-4\sqrt 7 }}}}{{\sqrt {12} }}\)
ОТВЕТ: 0,5.
\(\dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt[4]{{11-4\sqrt 7 }}}}{{\sqrt {12} }} = \dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt[4]{{7-4\sqrt 7 + 4}}}}{{\sqrt {12} }} = \) \( = \dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt[4]{{{{\left( {\sqrt 7 } \right)}^2}-2 \cdot 2 \cdot \sqrt 7 + {2^2}}}}}{{\sqrt {12} }} = \dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt[4]{{{{\left( {\sqrt 7 -2} \right)}^2}}}}}{{\sqrt {12} }} = \dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt {\left| {\sqrt 7 -2} \right|} }}{{\sqrt {12} }}.\) Так как \(\sqrt 7 > \sqrt 4 = 2\), то \(\sqrt 7 -2 > 0\) и \(\left| {\sqrt 7 -2} \right| = \sqrt 7 -2.\) \(\dfrac{{\sqrt {2 + \sqrt 7 } \cdot \sqrt {\left| {\sqrt 7 -2} \right|} }}{{\sqrt {12} }} = \dfrac{{\sqrt {\left( {2 + \sqrt 7 } \right) \cdot \left( {\sqrt 7 -2} \right)} }}{{\sqrt {12} }} = \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2}-{2^2}} }}{{\sqrt {12} }} = \dfrac{{\sqrt 3 }}{{\sqrt {12} }} = \sqrt {\dfrac{1}{4}} = \dfrac{1}{2} = 0,5.\) Ответ: 0,5.