Question

帮忙算一道数学题（要过程）

二次根号下，2007乘2008乘2009乘2010加1；根号外再减2008的平方

Answer 1

令a=2008
则2007乘2008乘2009乘2010加1
=(a-1)a(a+1)(a+2)+1
=[(a-1)(a+2)][a(a+1)]+1
=[(a²+a)-2](a²+a)+1
=(a²+a)²-2(a²+a)+1
=(a²+a-1)²

所以原式=√(a²+a-1)²-a²
=a²+a-1-a²
=a-1
=2008-1
=2007

Answer 2

原式=√(a²+a-1)²-a²
=a²+a-1-a²
=a-1
=2008-1
=2007

Answer 3

设t=2008
则2007乘2008乘2009乘2010加1
=(t-1)t(t+1)(t+2)+1
=[(t-1)(t+2)][t(t+1)]+1
=[(t2+t)-2](t2+t)+1
=(t2+t)2-2(t2+t)+1
=(t2+t-1)2

所以原式=√(t2+t-1)2-t2
=t2+t-1-t2
=t-1
=2008-1
=2007