Đáp án:

 a)x-y

b)$\frac{x^{2}-x+1}{2(x-1)(x+1)}$ 

Giải thích các bước giải:

 a)$\frac{x^{2}+y^{2}}{x-y}$ +$\frac{2xy}{y-x}$

=$\frac{x^{2}+y^{2}}{x-y}$-$\frac{2xy}{x-y}$                   

=$\frac{x^{2}+y^{2}-2xy}{x-y}$ 

=$\frac{x^{2}-2xy+y^{2}}{x-y}$

=$\frac{(x-y)^{2}}{x-y}$

=x-y

b)$\frac{5x-7}{2(x-1)}$ -$\frac{4x}{x^{2}-1}$ +$\frac{9-3x}{2(x-1)}$

=$\frac{5x-7}{2(x-1)}$-$\frac{4x}{(x-1)(x+1)}$ +$\frac{9-3x}{2(x-1)}$         Mẫu thức chung:2(x-1)(x+1)

=$\frac{5x^{2}-2x-7}{2(x-1)(x+1)}$ -$\frac{8x}{2(x-1)(x+1)}$ +$\frac{6x+9-3x^{2}}{2(x-1)(x+1)}$

=$\frac{5x^{2}-2x-7-8x+6x+9-3x^{2}}{2(x-1)(x+1)}$ 

=$\frac{2x^{2}-4x+2}{2(x-1)(x+1)}$ 

=$\frac{2(x^{2}-x+1)}{2(x-1)(x+1)}$

=$\frac{x^{2}-x+1}{(x-1)(x+1)}$ 

Đáp án:

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