thực hiện phép tính sau: (x^2+xy/x^3+x^2y+xy^2+y^3):(1/x-y-2xy/x^3-x^2y+xy^2-y^3) 11/08/2021 Bởi Anna thực hiện phép tính sau: (x^2+xy/x^3+x^2y+xy^2+y^3):(1/x-y-2xy/x^3-x^2y+xy^2-y^3)
Đáp án: $\begin{array}{l}\frac{{{x^2} + xy}}{{{x^3} + {x^2}y + x{y^2} + {y^3}}}:\left( {\frac{1}{{x – y}} – \frac{{2xy}}{{{x^3} – {x^2}y + x{y^2} – {y^3}}}} \right)\\ = \frac{{x\left( {x + y} \right)}}{{{x^2}\left( {x + y} \right) + {y^2}\left( {x + y} \right)}}:\left( {\frac{1}{{x – y}} – \frac{{2xy}}{{\left( {x – y} \right)\left( {{x^2} + {y^2}} \right)}}} \right)\\ = \frac{{x\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}:\left( {\frac{{{x^2} + {y^2} – 2xy}}{{\left( {x – y} \right)\left( {{x^2} + {y^2}} \right)}}} \right)\\ = \frac{x}{{{x^2} + {y^2}}}.\frac{{\left( {x – y} \right)\left( {{x^2} + {y^2}} \right)}}{{{{\left( {x – y} \right)}^2}}}\\ = \frac{x}{{x – y}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\frac{{{x^2} + xy}}{{{x^3} + {x^2}y + x{y^2} + {y^3}}}:\left( {\frac{1}{{x – y}} – \frac{{2xy}}{{{x^3} – {x^2}y + x{y^2} – {y^3}}}} \right)\\
= \frac{{x\left( {x + y} \right)}}{{{x^2}\left( {x + y} \right) + {y^2}\left( {x + y} \right)}}:\left( {\frac{1}{{x – y}} – \frac{{2xy}}{{\left( {x – y} \right)\left( {{x^2} + {y^2}} \right)}}} \right)\\
= \frac{{x\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}:\left( {\frac{{{x^2} + {y^2} – 2xy}}{{\left( {x – y} \right)\left( {{x^2} + {y^2}} \right)}}} \right)\\
= \frac{x}{{{x^2} + {y^2}}}.\frac{{\left( {x – y} \right)\left( {{x^2} + {y^2}} \right)}}{{{{\left( {x – y} \right)}^2}}}\\
= \frac{x}{{x – y}}
\end{array}$