解：设x=2001，原式变为：
√[(x-1)*x*(x+1)*(x+2)+1-(x-1)^2]
=√[x(x^2-1)(x+2)+1-(x-1)^2]
=√[x^4+2x^3-x^2-2x+1-x^2+2x-1]
=√[x^4+2x^3-2x^2]
=x√[(x+1)^2-3]
=2001√(2002^2-3)
=2001√4008001
由于4008001为质数，故原式=2001√4008001 。
设x=2001，原式变为：
根号[(x-1)*x*(x+1)*(x+2)+1-(x-1)^2]
=根号[x(x^2-1)(x+2)+1-(x-1)^2]
=根号[x^4+2x^3-x^2-2x+1-x^2+2x-1]
=根号[x^4+2x^3-2x^2]
=x根号[(x+1)^2-3]
=2001根号(2002^2-3)
=2001根号4008001
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