化简(x-2)/(x^2-9)[1+(2x-7)/(x^2-4x+4)]÷1/(x+3),并求x=2006时的值.

题目

化简(x-2)/(x^2-9)[1+(2x-7)/(x^2-4x+4)]÷1/(x+3),并求x=2006时的值.
初一分式的加减：化简(x-2)/(x^2-9)[1+(2x-7)/(x^2-4x+4)]÷1/(x+3),并求x=2006时的值

答案

(x-2)/(x^2-9)[1+(2x-7)/(x^2-4x+4)]÷1/(x+3)
=[(x-2)/(x+3)(x-3)][(x^2-4x+4+2x-7)/(x^2-4x+4)]×(x+3)
=[(x-2)/(x-3)][(x^2-2x-3)/(x^2-4x+4)]
=[(x-2)/(x-3)][(x-3)(x+2)/(x-2)^2]
=(x-2)(x-3)(x+2)/[(x-3)(x-2)^2]
=(x+2)/(x-2)
=(2006+2)/(2006-2)
=502/501