Упрощение алгебраических выражений. Задача 14math100admin44242025-03-26T20:48:19+03:00
Задача 14. Упростите выражение \(\left( {\dfrac{x}{{{x^2}-{y^2}}}-\dfrac{x}{{{{\left( {x-y} \right)}^2}}}} \right)\dfrac{{{y^2}-2xy + {x^2}}}{{2x}} + \dfrac{y}{{x + y}}\)
Решение
\(\left( {\dfrac{x}{{{x^2}-{y^2}}}-\dfrac{x}{{{{\left( {x-y} \right)}^2}}}} \right)\dfrac{{{y^2}-2xy + {x^2}}}{{2x}} + \dfrac{y}{{x + y}} = \)
\( = \left( {\dfrac{x}{{\left( {x-y} \right)\left( {x + y} \right)}}-\dfrac{x}{{{{\left( {x-y} \right)}^2}}}} \right) \cdot \dfrac{{{{\left( {x-y} \right)}^2}}}{{2x}} + \dfrac{y}{{x + y}} = \)
\( = \dfrac{{{x^2}-xy-{x^2}-xy}}{{{{\left( {x-y} \right)}^2}\left( {x + y} \right)}} \cdot \dfrac{{{{\left( {x-y} \right)}^2}}}{{2x}} + \dfrac{y}{{x + y}} = \dfrac{{-2xy}}{{\left( {x + y} \right)2x}} + \dfrac{y}{{x + y}} = -\dfrac{y}{{x + y}} + \dfrac{y}{{x + y}} = 0.\)
Ответ: 0.