Question

Answer 1

\(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)

\(=\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\\ =\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\\ =\dfrac{x^2-2xy+y^2}{x-y}\\ =\dfrac{\left(x-y\right)^2}{x-y}\\ =x-y\)

Answer 2

c) \(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)

=\(\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\) 

=\(\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\) 

=\(\dfrac{x^2-2xy+y^2}{x-y}\) 

=\(\dfrac{\left(x-y\right)^2}{x-y}\) 

=\(x-y\)