Задача 71. Упростите выражение    \(\dfrac{{\sqrt {1 + {{\left( {\dfrac{{{x^2}-1}}{{2x}}} \right)}^2}} }}{{\left( {{x^2} + 1} \right) \cdot {x^{-1}}}}\)

Ответ

ОТВЕТ: \(-\dfrac{1}{2},\) если   \(x \in \left( {-\infty ;\,0} \right);\)   \(\dfrac{1}{2},\) если   \(x \in \left( {0; + \infty } \right)\)

Решение

\(\dfrac{{\sqrt {1 + {{\left( {\dfrac{{{x^2}-1}}{{2x}}} \right)}^2}} }}{{\left( {{x^2} + 1} \right) \cdot {x^{-1}}}} = \dfrac{{\sqrt {1 + \dfrac{{{{\left( {{x^2}-1} \right)}^2}}}{{4{x^2}}}} }}{{\dfrac{{{x^2} + 1}}{x}}} = \sqrt {\dfrac{{4{x^2} + {x^4}-2{x^2} + 1}}{{4{x^2}}}}  \cdot \dfrac{x}{{{x^2} + 1}} = \)

\( = \dfrac{{\sqrt {{x^4} + 2{x^2} + 1} }}{{\sqrt {4{x^2}} }} \cdot \dfrac{x}{{{x^2} + 1}} = \dfrac{{\sqrt {{{\left( {{x^2} + 1} \right)}^2}} }}{{2\left| x \right|}} \cdot \dfrac{x}{{{x^2} + 1}} = \dfrac{{\left| {{x^2} + 1} \right|}}{{2\left| x \right|}} \cdot \dfrac{x}{{{x^2} + 1}} = \dfrac{{{x^2} + 1}}{{2\left| x \right|}} \cdot \dfrac{x}{{{x^2} + 1}} = \dfrac{x}{{2\left| x \right|}}.\)

Если  \(x\, \in \,\left( {-\infty ;0} \right)\), то  \(\dfrac{x}{{2\left| x \right|}} = -\dfrac{x}{{2x}} = -\dfrac{1}{2}\).

Если  \(x\, \in \,\left( {0;\infty } \right)\), то  \(\dfrac{x}{{2\left| x \right|}} = \dfrac{x}{{2x}} = \dfrac{1}{2}\).

Ответ:  \(-\dfrac{1}{2}\), если  \(x\, \in \,\left( {-\infty ;0} \right);\)   \(\dfrac{1}{2}\), если  \(x\, \in \,\left( {0;\infty } \right)\).