[$\frac{x^2-y^2}{xy}$ -$\frac{1}{x+y}$ ($\frac{x^2}{y}$- $\frac{y^2}{x}$)]:$\frac{x-y}{x}$ [

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1. $[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}(\dfrac{x^2}{y}-\dfrac{y^2}{x})]:\dfrac{x-y}{x}$

   $=[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}(\dfrac{x^3-y^3}{xy}].\dfrac{x}{x-y}$

   $=[\dfrac{x^2-y^2}{xy}-\dfrac{x^3-y^3}{xy(x+y)}].\dfrac{x}{x-y}$

   $=[\dfrac{(x^2-y^2)(x+y)-x^3+y^3}{xy(x+y)}].\dfrac{x}{x-y}$

   $=[\dfrac{x^3-xy^2+x^2y-y^3-x^3+y^3}{xy(x+y)}].\dfrac{x}{x-y}$

   $=\dfrac{x^2y-xy^2}{xy(x+y)}.\dfrac{x}{x-y}$

   $=\dfrac{xy(x-y)}{xy(x+y)}.\dfrac{x}{x-y}$

   $=\dfrac{x}{x+y}$

    

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2. Đáp án:

   \(\dfrac{x}{{x + y}}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   \left[ {\dfrac{{{x^2} – {y^2}}}{{xy}} – \dfrac{1}{{x + y}}.\dfrac{{{x^3} – {y^3}}}{{xy}}} \right]:\dfrac{{x – y}}{x}\\
    = \left( {\dfrac{{{x^2} – {y^2}}}{{xy}} – \dfrac{{{x^3} – {y^3}}}{{xy\left( {x + y} \right)}}} \right).\dfrac{x}{{x – y}}\\
    = \dfrac{{\left( {x + y} \right)\left( {{x^2} – {y^2}} \right) – {x^3} + {y^3}}}{{xy\left( {x + y} \right)}}.\dfrac{x}{{x – y}}\\
    = \dfrac{{{x^3} – x{y^2} + {x^2}y – {y^3} – {x^3} + {y^3}}}{{y\left( {x + y} \right)\left( {x – y} \right)}}\\
    = \dfrac{{xy\left( { – y + x} \right)}}{{y\left( {x + y} \right)\left( {x – y} \right)}}\\
    = \dfrac{{xy\left( {x – y} \right)}}{{y\left( {x + y} \right)\left( {x – y} \right)}}\\
    = \dfrac{x}{{x + y}}
   \end{array}\)

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