Toán Lớp 8: thực hiện phép cộng trừ phân thức
$\frac{1}{x-y}$ – $\frac{3xy}{x^3 – y^3}$ + $\frac{x-y}{x^2+xy+y^2}$
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Toán Lớp 8: thực hiện phép cộng trừ phân thức
$\frac{1}{x-y}$ – $\frac{3xy}{x^3 – y^3}$ + $\frac{x-y}{x^2+xy+y^2}$
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\dfrac{1}{{x – y}} – \dfrac{{3xy}}{{{x^3} – {y^3}}} – \dfrac{{x – y}}{{{x^2} + xy + {y^2}}}\\
= \dfrac{1}{{x – y}} – \dfrac{{3xy}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}} – \dfrac{{x – y}}{{{x^2} + xy + {y^2}}}\\
= \dfrac{{{x^2} + xy + {y^2}}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}} – \dfrac{{3xy}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}} – \dfrac{{{{\left( {x – y} \right)}^2}}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{{x^2} + xy + {y^2} – 3xy + {x^2} – 2xy + {y^2}}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{2{{\left( {x – y} \right)}^2}}}{{\left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{2\left( {x – y} \right)}}{{\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{2x – 2x}}{{\left( {{x^2} + xy + {y^2}} \right)}}
\end{array}\]