解1题
原式=[1-1/(1-x)]×[(1/x²)-1]÷[-(x+1)/x]
=[1+1/(x-1)]×[(1/x²)-(x²/x²)]×[-x/(x+1)]
=[(x-1+1)/(x-1)]×[(1-x²)/x²]×[-x/(x+1)]
=[x/(x-1)]×[-(x+1)(x-1)/x²]×[-x/(x+1)]
=1
解2题
原式=[(3-m)/(2m-4)]÷[(m+2)-5/(m-2)]
={(3-m)/[2(m-2)]}÷[(m+2)(m-2)/(m-2)-5/(m-2)]
={-(m-3)/[2(m-2)]}÷[(m²-4-5)/(m-2)]
={-(m-3)/[2(m-2)]}×{(m-2)/[(m+3)(m-3)]}
=-1/[2(m+3)]
=-1/[2×(1/2+3)]
=-1/(1+6)
=-1/7