∵3x+2y=13,2x-3y=0 ∴x=3,y=2 [(x^2+x-2)/(x^2+4x+4)]÷[(x^2+2x-3)/(x^2-4)]×[(x^2+6x+9)/(x^2-xy-2x+2y)] =[(x+2)(x-1)/(x+2)(x+2)]×[(x+2)(x-2)/(x+3)(x-1)]×[(x+3)(x+3)/(x-y)(x-2)] =(x+3)/(x-y) =(3+3)/(3-2) =6