若丨ab-2丨+(b-2)^2=0,求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1+……+(a+2004)(b+2004)分之1的值

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若丨ab-2丨+(b-2)^2=0,求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1+……+(a+2004)(b+2004)分之1的值

若丨ab-2丨+(b-2)^2=0,求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1+……+(a+2004)(b+2004)分之1的值
若丨ab-2丨+(b-2)^2=0,求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1+……+(a+2004)(b+2004)分之1的值

若丨ab-2丨+(b-2)^2=0,求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1+……+(a+2004)(b+2004)分之1的值
∵ 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
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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 ||
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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
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根据条件的得出a=1,b=2.
1+1/(2*3)+1/(3*4)+……+1/(2005*2006)
= 1+1/2-1/3+1/3-1/4+……+1/2005-1/2006
= 1+1/2-1/2006
=1504/1003

有两个非负数和为0得ab-2=0,b-1=0,所以a=2,b=1
结果就成了1/(1*2)+1/(2*3)+1/(3*4)+……+1/(2005*2006)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+……+(1/2005-1/2006)=1-1/2006=2005/2006