作业帮根号（2005×2006×2007×2008+1）-2006为什么会等于:令a=2006则根号内的=(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)-a²=a-1=200
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/11 04:43:17
![根号（2005×2006×2007×2008+1）-2006为什么会等于:令a=2006则根号内的=(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)-a²=a-1=200](/uploads/image/z/757202-50-2.jpg?t=%E6%A0%B9%E5%8F%B7%EF%BC%882005%C3%972006%C3%972007%C3%972008%2B1%EF%BC%89-2006%E4%B8%BA%E4%BB%80%E4%B9%88%E4%BC%9A%E7%AD%89%E4%BA%8E%3A%E4%BB%A4a%3D2006%E5%88%99%E6%A0%B9%E5%8F%B7%E5%86%85%E7%9A%84%3D%28a-1%29a%28a%2B1%29%28a%2B2%29%2B1%3D%5B%28a-1%29%28a%2B2%29%5D%5Ba%28a%2B1%29%5D%2B1%3D%5B%28a%26sup2%3B%2Ba%29-2%5D%28a%26sup2%3B%2Ba%29%2B1%3D%28a%26sup2%3B%2Ba%29%26sup2%3B-2%28a%26sup2%3B%2Ba%29%2B1%3D%28a%26sup2%3B%2Ba-1%29%26sup2%3B%E6%89%80%E4%BB%A5%E5%8E%9F%E5%BC%8F%3D%28a%26sup2%3B%2Ba-1%29-a%26sup2%3B%3Da-1%3D200)
根号（2005×2006×2007×2008+1）-2006为什么会等于:令a=2006则根号内的=(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)-a²=a-1=200
根号（2005×2006×2007×2008+1）-2006
为什么会等于:令a=2006
则根号内的=(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)-a²
=a-1
=2005
写出=(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
根号（2005×2006×2007×2008+1）-2006为什么会等于:令a=2006则根号内的=(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)-a²=a-1=200
这个啊,
因为a=2006,所以2005=a-1,2007=a+1,2008=a+2
所以2005×2006x2007x2008+1=(a-1)a(a+1)(a+2)+1
所以根号内2005×2006x2007x2008+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)-a²=a-1=2005
关键就是将原式里的数转换成用a代替的式子,再进行化简,使原式变得简单.
不要忽略“a=2006"
我这样写明白了吗?不明白再消息通知我吧
令a=2006
2005×2006×2007×2008+1=[(2006-1)(2006+2)][2006(2006+1)]+1==[(a-1)(a+2)][a(a+1)]+1
后面的知道了吧