a) Ta có
$\dfrac{x^2-xy}{3xy-3y^2} = \dfrac{x(x-y)}{3y(x-y)}$
$= \dfrac{x}{3y}$
b) Ta có
$\dfrac{x^6 + 2x^3 y^3 + y^6}{x^7 - xy^6} = \dfrac{(x^3)^2 + 2x^3 y^3 + (y^3)^2}{x(x^6 - y^6)}$
$=\dfrac{(x^3 + y^3)^2}{x[(x^3)^2 - (y^3)^2]}$
$= \dfrac{(x^3 + y^3)^2}{x(x^3 - y^3)(x^3 + y^3)}$
$= \dfrac{x^3 + y^3}{x(x^3 - y^3)}$