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원 식={[(x-2)/(x&\#178;+2x)]-[(x-1)/(x²+4x+4)]}÷[(x-4)/(x+2)]
={(x-2)/[x(x+2)]-[(x-1)/(x+2)²]}×[(x+2)/(x-4)]
=(x-2)/[x(x+2)]×[(x+2)/(x-4)]-[(x-1)/(x+2)²]×[(x+2)/(x-4)]
=(x-2)/[x(x-4)]-(x-1)/[(x+2)(x-4)]
=[(x-2)(x+2)]/[x(x+2)(x-4)]-[x(x-1)]/[x(x+2)(x-4)]
=[(x-2)(x+2)-x(x-1)]/[x(x+2)(x-4)]
=(x²-4-x²+x)/[x(x+2)(x-4)]
=(x-4)/[x(x+2)(x-4)]
=1/[x(x+2)
=1/(x²+2x)
x&\#178;+때문에2x-1=0
그래서 x&\#178;+2x=1
그래서.
오리지널=1/1=1