Đáp án:
\(\begin{array}{l}
a)\,\,\frac{{4x - y}}{{2x - 3y}}\\
b)\,\,\,\frac{2}{{x - y}}
\end{array}\)
Giải thích các bước giải:
$\begin{array}{l}
a)\,\,\,\frac{{4{x^2} - 5xy + {y^2}}}{{2{x^2} - 5xy + 3{y^2}}} = \frac{{4{x^2} - 4xy - xy + {y^2}}}{{2{x^2} - 2xy - 3xy + 3{y^2}}}\\
= \frac{{4x\left( {x - y} \right) - y\left( {x - y} \right)}}{{2x\left( {x - y} \right) - 3y\left( {x - y} \right)}} = \frac{{\left( {x - y} \right)\left( {4x - y} \right)}}{{\left( {x - y} \right)\left( {2x - 3y} \right)}} = \frac{{4x - y}}{{2x - 3y}}.\\
b)\,\,\frac{1}{{x - y}} - \frac{{3xy}}{{{y^3} - {x^3}}} + \frac{{x - y}}{{{x^2} + xy + {y^2}}}\\
= \frac{1}{{x - y}} + \frac{{3xy}}{{{x^3} - {y^3}}} + \frac{{x - y}}{{{x^2} + xy + {y^2}}}\\
= \frac{1}{{x - y}} + \frac{{3xy}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}} + \frac{{x - y}}{{{x^2} + xy + {y^2}}}\\
= \frac{{{x^2} + xy + {y^2} + 3xy + {{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \frac{{{x^2} + {y^2} + 4xy + {x^2} - 2xy + {y^2}}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \frac{{2{x^2} + 2{y^2} + 2xy}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}\\
= \frac{{2\left( {{x^2} + {y^2} + xy} \right)}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}} = \frac{2}{{x - y}}
\end{array}$
