∵2x=x+3 ∴x=3
1/(x+2)-[(x^2-4x+4)/(x^2-x)]/[x+1-3/(x-1)]
=1/(x+2)-[(x-2)^2/(x-1)x]/[(x^2-1-3)/(x-1)]
=1/(x+2)-[(x-2)^2/(x^3-4x)]
=[(x^3-4x)-(x+2)(x-2)^2]/[(x+2)(x^3-4x)]
=(x^2-4)(x-x+2)/(x^2-4)(x^2+2x)
=2/(x^2+2x)
x=3 时，原式=2/(3^2+2*3)=2/(9+6=2/15
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