Упрощение алгебраических выражений. Задача 28math100admin44242025-03-26T21:10:06+03:00
Задача 28. Упростите выражение \(\dfrac{1}{x}{\left( {\dfrac{{{y^2}-xy}}{{x + y}}} \right)^2}\left( {\dfrac{{x + y}}{{{{\left( {x-y} \right)}^2}}} + \dfrac{{x + y}}{{xy-{y^2}}}} \right) + \dfrac{x}{{x + y}}\)
Решение
\(\dfrac{1}{x}{\left( {\dfrac{{{y^2}-xy}}{{x + y}}} \right)^2}\left( {\dfrac{{x + y}}{{{{\left( {x-y} \right)}^2}}} + \dfrac{{x + y}}{{xy-{y^2}}}} \right) + \dfrac{x}{{x + y}} = \)
\( = \dfrac{1}{x} \cdot \dfrac{{{y^2} \cdot {{\left( {y-x} \right)}^2} \cdot \left( {x + y} \right)}}{{{{\left( {x + y} \right)}^2}}} \cdot \left( {\dfrac{1}{{{{\left( {x-y} \right)}^2}}} + \dfrac{1}{{y\left( {x-y} \right)}}} \right) + \dfrac{x}{{x + y}} = \)
\( = \dfrac{1}{x} \cdot \dfrac{{{y^2} \cdot {{\left( {x-y} \right)}^2}}}{{x + y}} \cdot \dfrac{{y + x-y}}{{y{{\left( {x-y} \right)}^2}}} + \dfrac{x}{{x + y}} = \)
\( = \dfrac{1}{x} \cdot \dfrac{{y \cdot x}}{{x + y}} + \dfrac{x}{{x + y}} = \dfrac{y}{{x + y}} + \dfrac{x}{{x + y}} = \dfrac{{x + y}}{{x + y}} = 1.\)
Ответ: 1.