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+2
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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+2](/uploads/image/z/14114411-35-1.jpg?t=%E8%8B%A5%E4%B8%A8ab-2%E4%B8%A8%2B%EF%BC%88b-2%EF%BC%89%5E2%3D0%2C%E6%B1%82ab%E5%88%86%E4%B9%8B1%2B%EF%BC%88a%2B1%EF%BC%89%EF%BC%88b%2B1%EF%BC%89%E5%88%86%E4%B9%8B1%2B%EF%BC%88a%2B2%EF%BC%89%EF%BC%88b%2B2%EF%BC%89%E5%88%86%E4%B9%8B1%2B%E2%80%A6%E2%80%A6%2B%EF%BC%88a%2B2)
∵ 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
-------------------------------------------------------------------------------------
(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
-------------------------------------------------------------------------------------
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