#andy

\[\begin{array}{l}
a)\dfrac{{{x^2} + {y^2}}}{{x - y}} + \dfrac{{2xy}}{{y - x}}\\
 = \dfrac{{{x^2} + {y^2}}}{{x - y}} - \dfrac{{2xy}}{{x - y}}\\
 = \dfrac{{{x^2} + {y^2} - 2xy}}{{x - y}}\\
 = \dfrac{{{{\left( {x - y} \right)}^2}}}{{x - y}}\\
 = x - y\\
b)\dfrac{{5x - 7}}{{2\left( {x - 1} \right)}} - \dfrac{{4x}}{{{x^2} - 1}} + \dfrac{{9 - 3x}}{{2\left( {x - 1} \right)}}\\
 = \dfrac{{5x - 7}}{{2\left( {x - 1} \right)}} - \dfrac{{4x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{{9 - 3x}}{{2\left( {x - 1} \right)}}\\
 = \dfrac{{\left( {5x - 7} \right)\left( {x + 1} \right) - 8x + \left( {9 - 3x} \right)\left( {x + 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{5{x^2} - 2x - 7 - 8x + \left( {9 - 3x} \right)\left( {x + 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2{x^2} - 4x + 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2\left( {{x^2} - 2x + 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2{{\left( {x - 1} \right)}^2}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{x - 1}}{{x + 1}}
\end{array}\]

Mong cho lời giải hay nhất ❤️⭐️