求一因式分解题答案
来源:百度知道 编辑:UC知道 时间:2024/05/14 05:41:10
要过程,谢谢
本人电脑水平不好,打不了平方,还望大家看题仔细点
谢谢
1-1/(2的平方))*(1-1/(3的平方)).....(1-1/(2006的平方)
=[1+1/2][1-1/2][1+1/3][1-1/3]....[1+1/2006][1-1/2006]
=3/2*1/2*4/3*2/3....2007/2006*2005/2006
=1/2*2007/2006
=2007/4012
a^2-b^2=[a+b][a-b]
解:
应用平方差公式得:
原式= (1-1/2)(1+1/2)*(1/1/3)(1+1/3)*...*(1-1/2006)(1+1/2006)
= [(1-1/2)(1-1/3)...(1-1/2006)]* [(1+1/2)(1+1/3)...(1+1/2006)]
= (1/2 *2/3 *3/4* ...*2005/2006)* (3/2 *4/3 *5/4 *...*2007/2006)
= 1/2006 * 2007/2
= 2007/4012
利用a^2-b^2=(a+b)*(a-b)
(1-1/(2的平方))*(1-1/(3的平方)).....(1-1/(2006的平方)
=(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*...*(1+1/2006)*(1-1/2006)
=[(1+1/2)*(1+1/3)*...*(1+1/2006)]*[(1-1/2)*(1-1/3)*...*(1-1/2006)]
=[(3/2)*(4/3)*(5/4)*...*(2007/2006)]*[(1/2)*(2/3)*3/4)*...*(2005/2006)]
=(2007/2)*(1/2006)
=2007/4012
解:(1-1/(2的平方))*(1-1/(3的平方)).....(1-1/(2006的平方)
=(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*...*(1+1/2006)*(1-1/2006)