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Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)
\(=\dfrac{2006}{2007}\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)
Áp dụng Bất đẳng thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(\dfrac{2006^{2006}+1}{2006^{2007}+1}< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}=\dfrac{2006^{2006}+2006}{2006^{2007}+2006}=\dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}=\dfrac{2006^{2005}+1}{2006^{2006}+1}\)
\(\Leftrightarrow\dfrac{2006^{2006}+1}{2006^{2007}+1}< \dfrac{2006^{2005}+1}{2006^{2006}+1}\)
\(\dfrac{1}{38}>\dfrac{1}{40}>\dfrac{1}{42}>...>\dfrac{1}{50}\)
=>\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+\dfrac{1}{44}+\dfrac{1}{46}+\dfrac{1}{48}+\dfrac{1}{50}< 7\cdot\dfrac{1}{38}=\dfrac{7}{38}< 1\)
Vậy tổng trên bé hơn 1
A=-1-3-5-...-2017
=-(1+3+5+...+2017)
Xét tổng B=1+3+5+...+2017
Tổng B có:(2017-1):2+1=1009(số hạng)
Tổng B=\(\dfrac{\left(2017+1\right)\cdot1009}{2}=1009\cdot1009=1018081\)
=>A=-B=-1018081
Ta có:
\(2006A=\dfrac{2006^{2007}+2016}{2006^{2007}+1}=1+\dfrac{2005}{2006^{2007}+1}\)
\(2006B=\dfrac{2006^{2006}+2006}{2006^{2006}+1}=1+\dfrac{2005}{2006^{2006}+1}\)
Do \(\dfrac{2005}{2006^{2006}+1}>\dfrac{2005}{2006^{2007}+1}\Rightarrow1+\dfrac{2005}{2006^{2006}+1}>1+\dfrac{2005}{2006^{2007}+1}\)
\(\Rightarrow2006A< 2006B\Rightarrow A< B\)
Mình sẽ giải cách ngắn hơn cách bạn đạt nha:
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}< 1\)
\(A< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}\Rightarrow A< \dfrac{2006^{2006}+2006}{2006^{2007}+2006}\Rightarrow A< \dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}\Rightarrow A< \dfrac{2006^{2005}+1}{2006^{2006}+1}=B\)\(A< B\)
=>B=\(\dfrac{1}{4.4}+\dfrac{1}{6.6}+\dfrac{1}{8.8}+...+\dfrac{1}{2006.2006}\)
=>B<\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2005.2007}\)
=>B<\(\dfrac{2}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2005.2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2005.2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{2005}-\dfrac{1}{2005}-\dfrac{1}{200}\right)\)(xin lỗi, đoạn cuối (chỗ 200 í )là 2007 nhá
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{2007}\right)\)
=>B<\(\dfrac{1}{2}.\dfrac{668}{2007}\)
=>B<\(\dfrac{1.668}{2.2007}\)
=>B<\(\dfrac{1.668:2}{2.2007:2}\)
=>B<\(\dfrac{334}{2007}\)
Tick cho tôi nha :D
Lời giải:
Không biết đây có phải cách tối ưu nhất hay không nhưng tạm thời giờ mình nghĩ theo hướng này:
\(P=\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\)
Ghép cặp:
\(\frac{1}{2006}+\frac{1}{2014}=\frac{4020}{2006.2014}=\frac{2.2010}{(2010-4)(2010+4)}=\frac{2.2010}{2010^2-4^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2007}+\frac{1}{2013}=\frac{4020}{2007.2013}=\frac{2.2010}{(2010-3)(2010+3)}=\frac{2.2010}{2010^2-3^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2008}+\frac{1}{2012}=\frac{4020}{2008.2012}=\frac{2.2010}{(2010-2)(2010+2)}=\frac{2.2010}{2010^2-2^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2009}+\frac{1}{2011}=\frac{4020}{2009.2011}=\frac{2.2010}{(2010-1)(2010+1)}=\frac{2.2010}{2010^2-1^2}>\frac{2.2010}{2010^2}=\frac{2}{2010}\)
\(\frac{1}{2005}> \frac{1}{2010}\)
\(\frac{1}{2010}=\frac{1}{2010}\)
Cộng tất cả các kết quả trên lại:
\(P> \frac{2}{2010}+\frac{2}{2010}+\frac{2}{2010}+\frac{2}{2010}+\frac{1}{2010}+\frac{1}{2010}\)
\(\Leftrightarrow P> \frac{10}{2010}=\frac{1}{201}\Rightarrow \frac{1}{P}< 201\)
ta có
1/2005>1/2014
1/2006>1/2014
...
1/2014=1/2014
=> 1/2005+1/2005+1/2006+1/2007+...+<1/2014.10
=>1/2005+1/2005+...+1/2014<10.1/2014<10.1/2010=1/201
=>P<1/201
=>1/P<201