已知x^2+5*x-2004=0，求分式((x-2)^3-(x-1)^2+1) /(x-2)的值。
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x^2-5x-2004=0
x^2-5x=2004
[(x-2)^3-(x-1)^2+1] /(x-2)
=[(x-2)^3+1-(x-1)^2] /(x-2)
=[(x-2)^3+(1-x+1)(1+x-1)] /(x-2)
=[(x-2)^3+x(2-x)] /(x-2)
=[(x-2)^3-x(x-2)] /(x-2)
=(x-2)[(x-2)^2-x] /(x-2)
=(x-2)^2-x
=x^2-4x+4-x
=x^2-5x+4
=2004-4
=2000
x^2-5x-2004=0
x^2-5x=2004
[(x-2)^3-(x-1)^2+1] /(x-2)
=[(x-2)^3+1-(x-1)^2] /(x-2)
=[(x-2)^3+(1-x+1)(1+x-1)] /(x-2)
=[(x-2)^3+x(2-x)] /(x-2)
=[(x-2)^3-x(x-2)] /(x-2)
=(x-2)[(x-2)^2-x] /(x-2)
=(x-2)^2-x
=x^2-4x+4-x
=x^2-5x+4
=2004-4
=2000
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