解(1) 　　原式=1/(m²-4)+1/(m+2)+1/(2-m) 　　=1/[(m+2)(m-2)]+1/(m+2)-1/(m-2) 　　=1/[(m+2)(m-2)]+(m-2)/[(m+2)(m-2)]-(m+2)/[(m+2)(m-2)] 　　=[1+(m-2)-(m+2)]/[(m+2)(m-2)] 　　=(1+m-2-m-2)/[(m+2)(m-2)] 　　=-3/[(m+2)(m-2)] 　　解(2) 　　原式=[(x²-y²)/(x+y)]÷[2-(x²+y²)/(xy)] 　　=[(x+y(x-y)/(x+y)]÷[(2xy/xy)-(x²+y²)/(xy)] 　　=(x-y)÷[-(x²-2xy+y²)/(xy)] 　　=(x-y)×[-xy/(x²-2xy+y²)] 　　=(x-y)×[-xy/(x-y)²] 　　=-xy/(x-y) 　　解(3) 　　原式=[x/(x-2)-x/(x+2)]÷[4x/(2-x)] 　　={x(x+2)]/[(x+2)(x-2)]-x(x-2)/[(x+2)(x-2)]}×[-(x-2)/(4x)] 　　={[x(x+2)-x(x-2)]/[(x+2)(x-2)}×[-(x-2)/(4x)] 　　={(x²+2x-x²+2x)/[(x+2)(x-2)]}×[-(x-2)/(4x)] 　　={(4x)/[(x+2)(x-2)]}×[-(x-2)/(4x)] 　　=-1/(x+2) 　　解(4) 　　原式=[(a²+a)/(a-1)]÷[a-a/(a-1)] 　　=[a(a+1)/(a-1)]÷[a(a-1)/(a-1)-a/(a-1)] 　　=[a(a+1)/(a-1)]÷{[a(a-1)-a]/(a-1)] 　　=[a(a+1)/(a-1)]÷[(a²-a-a)/(a-1)] 　　=[a(a+1)/(a-1)]×[(a-1)/(a²-2a)] 　　=[a(a+1)/(a-1)]×(a-1)/[a(a-2)] 　　=(a+1)/(a-2) 　　解(5) 　　原式=[x/(x²-1)]×[(x²+2x+1)/(x-2)(x+1)]÷[2x/(x-1)] 　　=[x/(x+1)(x-1)]×[(x+1)²/(x-2)(x+1)]×[(x-1)/(2x)] 　　=1/[2(x-2)] 　　解(6) 　　原式=1/(m-3)-[1/(m²-9)]÷[(m-1)/(6+2m)] 　　=1/(m-3)-[1/(m+3)(m-3)]×[2(m+3)/(m-1)] 　　=1/(m-3)-[2/(m-3)(m-1)] 　　=(m-1)/(m-3)(m-1)-[2/(m-3)(m-1)] 　　=[(m-1)-2]/[(m-3)(m-1)] 　　=(m-3)/[(m-3)(m-1)] 　　=1/(m-1) 　　解(7) 　　原式=[(x-3)/(x²-1)]÷[(x²-2x-3)/(x²+2x+1)]+1/(x-1) 　　=[(x-3)/(x+1)(x-1)]÷[(x-3)(x+1)/(x+1)²]+1/(x-1) 　　=[(x-3)/(x+1)(x-1)]×[(x+1)/(x-3)]+1/(x-1) 　　=1/(x-1)+1/(x-1) 　　=2/(x-1)