tim x : 1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)-1/1/2010
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\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\).
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(=-\frac{1}{x+3}=\frac{1}{2010}\)
\(x=2010-\left(-3\right)=2013\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{2005}{2010}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{401}{402}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{401}{402}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{401}{402}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{402}\)
\(\Leftrightarrow x+1=402\Rightarrow x=401\)
Xét: \(1+2+3+.....+2010\) là dãy số tự nhiên cách đều
Tổng bằng:
\(\dfrac{\left(2010+1\right)\times\left[\left(2010-1\right):1+1\right]}{2}=\dfrac{2011\times2010}{2}=\dfrac{4042110}{2}=2021055\)(1)
A=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
A=\(\frac{2009}{2010}.\frac{2008}{2010}...0.\frac{-1}{2010}\)
A=0
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{-1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow-\left(x-3\right)=2010\)
\(\Rightarrow x=-2013\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow x=2007\)
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