Упрощение алгебраических выражений. Задача 35math100admin44242025-03-26T21:25:13+03:00
Задача 35. Упростите выражение \(\left( {\left( {a-b} \right)\sqrt {\dfrac{{a + b}}{{a-b}}} + a-b} \right)\dfrac{1}{{6b}}\left( {\sqrt {\dfrac{{a + b}}{{a-b}}} -1} \right)\)
Решение
\(\left( {\left( {a-b} \right)\sqrt {\dfrac{{a + b}}{{a-b}}} + a-b} \right)\dfrac{1}{{6b}}\left( {\sqrt {\dfrac{{a + b}}{{a-b}}} -1} \right) = \left( {a-b} \right)\left( {\dfrac{{\sqrt {a + b} }}{{\sqrt {a-b} }} + 1} \right) \cdot \dfrac{1}{{6b}} \cdot \left( {\dfrac{{\sqrt {a + b} -\sqrt {a-b} }}{{\sqrt {a-b} }}} \right) = \)
\( = \dfrac{{a-b}}{{6b}} \cdot \dfrac{{\sqrt {a + b} + \sqrt {a-b} }}{{\sqrt {a-b} }} \cdot \dfrac{{\sqrt {a + b} -\sqrt {a-b} }}{{\sqrt {a-b} }} = \dfrac{{a-b}}{{6b}} \cdot \dfrac{{a + b-a + b}}{{{{\left( {\sqrt {a-b} } \right)}^2}}} = \dfrac{{a-b}}{{6b}} \cdot \dfrac{{2b}}{{a-b}} = \frac{1}{3}.\)
Ответ: \(\dfrac{1}{3}.\)