Упрощение алгебраических выражений. Задача 69

ОТВЕТ: \(\dfrac{1}{{\sqrt {b + 2} }}\).

\(\dfrac{{\sqrt {2b + 2\sqrt {{b^2}-4} } }}{{\sqrt {{b^2}-4}  + b + 2}}.\)

\(b\, \in \,\left[ {2; + \infty } \right).\)

\(\dfrac{{\sqrt {2b + 2\sqrt {{b^2}-4} } }}{{\sqrt {{b^2}-4}  + b + 2}} = \dfrac{{\sqrt {b + 2 + 2\sqrt {b-2} \sqrt {b + 2}  + b-2} }}{{\sqrt {b-2}  \cdot \sqrt {b + 2}  + {{\left( {\sqrt {b + 2} } \right)}^2}}} = \)

\( = \dfrac{{\sqrt {{{\left( {\sqrt {b + 2} } \right)}^2} + 2\sqrt {b-2} \sqrt {b + 2}  + {{\left( {\sqrt {b-2} } \right)}^2}} }}{{\sqrt {b + 2}  \cdot \left( {\sqrt {b-2}  + \sqrt {b + 2} } \right)}} = \dfrac{{\sqrt {{{\left( {\sqrt {b + 2}  + \sqrt {b-2} } \right)}^2}} }}{{\sqrt {b + 2}  \cdot \left( {\sqrt {b-2}  + \sqrt {b + 2} } \right)}} = \)

\( = \dfrac{{\sqrt {b + 2}  + \sqrt {b-2} }}{{\sqrt {b + 2}  \cdot \left( {\sqrt {b-2}  + \sqrt b  + 2} \right)}} = \dfrac{1}{{\sqrt {b + 2} }}.\)

Ответ:  \(\dfrac{1}{{\sqrt {b + 2} }}.\)