根号下【2000*2001*2002*2003+1-2000*2000】怎样开,
问题描述:
根号下【2000*2001*2002*2003+1-2000*2000】怎样开,
答
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
答
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...