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Xét N :
N = \(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+...+\(\frac{1}{2009.2009}\)+\(\frac{1}{2010.2010}\)
Ta có :
\(\frac{1}{2.2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3.3}\)< \(\frac{1}{2.3}\)
...
\(\frac{1}{2009.2009}\)<\(\frac{1}{2008.2009}\)
\(\frac{1}{2010.2010}\)<\(\frac{1}{2019.2010}\)
Cộng vế theo vế của các bất đẳng thức trên , ta có :
\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+...+\(\frac{1}{2009.2009}\)+\(\frac{1}{2010.2010}\) < \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2008.2009}\)+\(\frac{1}{2019.2010}\)
=> N < 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
=> N < 1 - \(\frac{1}{2010}\)<1
=> N < 1
\(Cm:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< 1\)
Có : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2008\cdot2009}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{1}{1}-\frac{1}{2009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{2008}{2009}\left(1\right)\)
Mà \(\frac{2008}{2009}< 1\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< \frac{2008}{2009}< 1\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}< 1\left(đpcm\right)\)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2009^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2008.2009}\\ =\frac{1}{1}-\frac{1}{2009}< 1\left(\text{đ}pcm\right)\)
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2009^2}+\frac{1}{2010^2}>1\)
=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2009^2}+\frac{1}{2010^2}>\frac{ }{ }\)\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2008.2009}+\frac{1}{2009.2010}\)
=\(\frac{1}{1}-\frac{1}{2010}=\frac{2010}{2010}-\frac{1}{2010}\)=\(\frac{2010}{2010}>\frac{1}{2010}=1>\frac{1}{2010}\)
Vậy \(1>\frac{1}{2010}\)
Bạn ơi sai đề nhé
Bài 1 :
\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\cdot\frac{24}{50}=1\)
\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)
#Louis
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{12}{25}x=1\)
Đến đây dễ rồi :)))
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