Cho A = \(\frac{2000}{2001}+\frac{2001}{2002}+\frac{2002}{2003}+\frac{2003}{2004}+\frac{2005}{2006}+\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)
Hãy so sánh tổng các phân số trong A và so sánh với 15.
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
So sành \(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}\)với 8
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
Ta có:
2002/2001=1+1/2001
2003/2002=1+1/2002
2004/2003= 1+ 1/2003
2005/2004= 1+ 1/2004
2006/2005=1+ 1/2005
2007/2006= 1+ 1/2006
2008/2007=1 + 1/2007.
2009/2008=1+ 1/2008.
=> 2002/2001+2003/2002+2004?2003+2005/2004+2006/2005+ 2007/2006+ 2008/2007+ 2009/2008= 1+1+1+1+1+1+1+1+1/2001+1/2002+1/2003+1/2004+1/2005+1/2006+1/2007+1/2008>8.
Nhớ k đúng cho mình nha!! Thanks!!!
\(S=\frac{1}{\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+...+\frac{1}{2009}+\frac{1}{2010}}\)
Tìm số nguyên lớn nhất bé hơn \(S.\)
Có: \(S\le\frac{1}{\frac{\left(1+1+1+...+1\right)^2}{2001+2002+2003+...+2010}}=\frac{1}{\frac{10^2}{20055}}=\frac{4011}{20}=200,55\)
Do \(\frac{1}{2001}\ne\frac{1}{2002}\ne\frac{1}{2003}\ne...\ne\frac{1}{2010}\) nên dấu "=" không xảy ra \(\Rightarrow\)\(S< 200,55\) (1)
Lại có: \(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+...+\frac{1}{2010}< \frac{1}{2001}+\frac{1}{2001}+...+\frac{1}{2001}=\frac{10}{2001}\)
\(\Rightarrow\)\(S>\frac{2001}{10}=200,1\) (2)
(1) và (2) suy ra \(200,1< S< 200,55\)\(\Rightarrow\) số nguyên lớn nhất bé hơn S là 200
PS: sai chỗ nào mn chỉ ạ :3
So sánh : \(A=\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}vàB=\frac{2008+2009+2010}{2009+2010+2011}\)
BT1: Tính:
\(\frac{1}{1+\frac{2009}{2009}+\frac{2009}{2010}}\) + \(\frac{1}{1+\frac{2010}{2009}+\frac{2010}{2011}}\) + \(\frac{1}{1+\frac{2011}{2009}+\frac{2011}{2010}}\)
CMR\(\frac{a+2009}{a-2009}=\frac{b+2010}{b-2010}thi\frac{a}{2009}=\frac{b}{2010}\)
+ \(\frac{a}{2009}=\frac{b}{2010}\Leftrightarrow2010a=2009b.\)(1)
+ \(\frac{a+2009}{a-2009}=\frac{b+2010}{b-2010}\Rightarrow\left(a+2009\right)\left(b-2010\right)=\left(a-2009\right)\left(b+2010\right)\)
\(\Rightarrow ab-2010a+2009b-2009.2010=ab+2010a-2009b-2009.2010\)
\(\Leftrightarrow2.2009.b=2.2010.a\Leftrightarrow2010a=2009b\)(2)
Từ (1) và (2) => dpcm
\(B=\frac{\frac{2008}{2011}+\frac{2009}{2010}+\frac{2010}{2009}+\frac{2011}{2008}+\frac{2012}{503}}{\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}}\)
Cho \(\frac{2010\cdot c-2011\cdot b}{2009}=\frac{2011\cdot a-2009\cdot c}{2010}=\frac{2009\cdot b-2010\cdot c}{2011}\)
C/m \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}\)
So sánh : A=\(\frac{2008}{2009}\)+\(\frac{2009}{2010}\)+\(\frac{2010}{2011}\)và B=\(\frac{2008+2009+2010}{2009+2010+2011}\)
\(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
\(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}=\frac{2009}{2009+2010+2011}=\frac{2010}{2009+2010+2011}\)
\(< A=\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}\)
