第2个回答 2019-04-07
解:
∵
丨ab-2丨=0
(b-2)²=0
∴{
ab-2=0
①
{
b-2=0
②
解二元一次方程组得:
{a=1
{b=2
又∵1/(a+1)(b+1)=1/(1×2)=(1-1/2)
∴1/(1×2)+1/(2×3)+1/(3×4)+……+1/(2005×2006)
(化简得:)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2005-1/2006)
(去括号得:)
=1-1/2006
=2005/2006
答:1/(1×2)+1/(2×3)+……+1/(2005×2006)的值是2005/2006
-------------------------------------------------------------------------------------
解析:
因为绝对值里面的数不是正数就是“0”,第二个式子因为有“²”所以它的数也不是正数就是“0”。又因为只有0+0=0。所以:
∴{
ab-2=0
①
||
又因为ab-2=0等于ab=2
||再把a=1
b=2一带
{
b-2=0
②
||
b-2=0等于b=2
||
-----------------------
1/(a+1)(b+1)=1/(1×2)=(1-1/2)
=1/(1×2)+1/(2×3)+1/(3×4)+……+1/(2005×2006)
(化简得:)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2005-1/2006)
(去括号得:)
=1-1/2006
=2005/2006
-------------------------------------------------------------------------------------