x²-5x-2000=0
x²-5x=2000
于是
[(x-2)³-(x-1)²+1]/(x-2)
=[(x-2)³-(x²-2x+1)+1]/(x-2)
=[(x-2)³-x²+2x]/(x-2)
=[(x-2)³-x(x-2)]/(x-2)
=(x-2)[(x-2)²-x]/(x-2)
=(x-2)²-x
=x²-4x+4-x
=x²-5x+4
=2000+4
=2004
x²-5x-2000=0
x²-5x=2000
于是
[(x-2)³-(x-1)²+1]/(x-2)
=[(x-2)³-(x²-2x+1)+1]/(x-2)
=[(x-2)³-x²+2x]/(x-2)
=[(x-2)³-x(x-2)]/(x-2)
=(x-2)[(x-2)²-x]/(x-2)
=(x-2)²-x
=x²-4x+4-x
=x²-5x+4
=2000+4
=2004