\ \left \{ {{x-y=2} \atop {xy\left(x-y\right)=6}} \right. ==>\left \{ {{x-y=2} \atop {xy=3}} \right. \\ x=y+2;\\ (y+2)y=3==> \ \ \ y^2+2y-3=0;\\ D=4+4\cdot3=16;\\ y_1=\frac{-2-4}{2}=-3; x_1=-3+2=-1;\\ y_2=\frac{-2+4}{2}=1; x_2=1+2=3;\\ " alt=" \left \{ {{x-y=2} \atop {x^2y-xy^2=6}} \right. ==>\ \left \{ {{x-y=2} \atop {xy\left(x-y\right)=6}} \right. ==>\left \{ {{x-y=2} \atop {xy=3}} \right. \\ x=y+2;\\ (y+2)y=3==> \ \ \ y^2+2y-3=0;\\ D=4+4\cdot3=16;\\ y_1=\frac{-2-4}{2}=-3; x_1=-3+2=-1;\\ y_2=\frac{-2+4}{2}=1; x_2=1+2=3;\\ " align="absmiddle" class="latex-formula">
имеем два решения, гаибольшее значение возможно при паре (3;1)
3+1=4
Ответ:4