Упрощение алгебраических выражений. Задача 44math100admin44242025-03-26T21:44:28+03:00
Задача 44. Упростите выражение \(\left( {\dfrac{{x + \sqrt {{x^2}-4} + 2}}{{x-\sqrt {{x^2}-4} + 2}} + \dfrac{{2-\sqrt {{x^2}-4} + x}}{{2 + \sqrt {{x^2}-4} + x}}} \right)\,{x^{-1}}\)
Ответ
ОТВЕТ: 1, если \(x \in \left( {-\infty ;-2} \right) \cup \left[ {\,2;\, + \infty } \right).\)
Решение
\(\left( {\dfrac{{x + \sqrt {{x^2}-4} + 2}}{{x-\sqrt {{x^2}-4} + 2}} + \dfrac{{2-\sqrt {{x^2}-4} + x}}{{2 + \sqrt {{x^2}-4} + x}}} \right) \cdot {x^{-1}} = \)
\( = \left( {\dfrac{{{{\left( {\sqrt {x + 2} } \right)}^2} + \sqrt {x + 2} \sqrt {x-2} }}{{{{\left( {\sqrt {x + 2} } \right)}^2}-\sqrt {x + 2} \sqrt {x-2} }} + \dfrac{{{{\left( {\sqrt {x + 2} } \right)}^2}-\sqrt {x-2} \sqrt {x + 2} }}{{{{\left( {\sqrt {x + 2} } \right)}^2} + \sqrt {x-2} \sqrt {x + 2} }}} \right) \cdot \dfrac{1}{x} = \)
\( = \left( {\dfrac{{\sqrt {x + 2} \left( {\sqrt {x + 2} + \sqrt {x-2} } \right)}}{{\sqrt {x + 2} \left( {\sqrt {x + 2} -\sqrt {x-2} } \right)}} + \dfrac{{\sqrt {x + 2} \left( {\sqrt {x + 2} -\sqrt {x-2} } \right)}}{{\sqrt {x + 2} \left( {\sqrt {x + 2} + \sqrt {x-2} } \right)}}} \right) \cdot \dfrac{1}{x} = \)
\( = \left( {\dfrac{{\sqrt {x + 2} + \sqrt {x-2} }}{{\sqrt {x + 2} -\sqrt {x-2} }} + \dfrac{{\sqrt {x + 2} -\sqrt {x-2} }}{{\sqrt {x + 2} + \sqrt {x-2} }}} \right)\dfrac{1}{x} = \)
\( = \dfrac{{x + 2 + 2\sqrt {x + 2} \sqrt {x-2} + x-2 + x + 2-2\sqrt {x + 2} \sqrt {x-2} + x-2}}{{x + 2-x + 2}} \cdot \dfrac{1}{x} = \)
\( = \dfrac{{4x}}{4} \cdot \dfrac{1}{x} = 1\), если \(x\, \in \,\left( {-\infty ;-2} \right) \cup \left[ {2;\infty } \right).\)
Ответ: 1, если \(x\, \in \,\left( {-\infty ;-2} \right) \cup \left[ {2;\infty } \right).\)