So sánh :
A = \(\dfrac{2006}{2007}\) - \(\dfrac{2007}{2008}\) + \(\dfrac{2008}{2009}\) - \(\dfrac{2009}{2010}\)
B = \(-\dfrac{1}{2006.2007}\) - \(\dfrac{1}{2008.2009}\)
help me ~ mai mik nộp rồi
so sánh 2 số hữu tỉ:
A=\(\dfrac{2006}{2007}\)-\(\dfrac{2007}{2008}\)+\(\dfrac{2008}{2009}\)-\(\dfrac{2009}{2010}\)
B=\(\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
mình đang cần gấp lắm nhé, mọi người giúp mình nhé.
hơi giống bài mk mà bài mk là -1/2006*2007 cơ :)
Vào link sau để xem câu trả lời vì hai bài giống nhau này
https://hoc24.vn/hoi-dap/question/655364.html
So sánh các số hữu tỉ:
a, x=\(\dfrac{-24}{35}\) ; y=\(\dfrac{-19}{30}\)
b, A= \(\dfrac{2006}{2007}\) - \(\dfrac{2007}{2008}\) + \(\dfrac{2008}{2009}\) - \(\dfrac{2009}{2010}\)
B= \(\dfrac{-1}{2006.2007}\) - \(\dfrac{1}{2008.2009}\)
a) Ta có:
\(-\dfrac{24}{35}< -\dfrac{24}{30}< -\dfrac{19}{30}\)
\(\Rightarrow x< y\)
b) Ta có:
\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
Ta lại có:
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
=> Dễ dàng thấy A > B
So sánh A và B biết: a=2006/2007-2007/2008-2008/2009-2009/2010 và B= -1/2006.2007-1/2008.2009
so sánh 2006/2007-2007/2008+2008/2009-2009/2010 và -1/2006.2007+-1/2008.2009
So sánh
M=2006/2007 - 2007/2008 + 2008/2009 - 2009/2010
N= -1/2006.2007 - 1/2008.2009
Cho A=\(\dfrac{2006}{2007}+\dfrac{2007}{2008}+\dfrac{2008}{2009}+\dfrac{2009}{2006}\) .Hãy so sánh số đó với 4
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)
\(A=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)+\left(1+\frac{3}{2006}\right)\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)
\(A=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
\(A=4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
Ta có: \(\left\{{}\begin{matrix}\frac{1}{2007}< \frac{1}{2006}\\\frac{1}{2008}< \frac{1}{2006}\\\frac{1}{2009}< \frac{1}{2006}\end{matrix}\right.\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}< \frac{1}{2006}+\frac{1}{2006}+\frac{1}{2006}=\frac{3}{2006}\)
\(\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}< 0\)
\(\Rightarrow4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)>4\)
hay \(A>4\)
\(\text{Vậy A>4}\)
So sánh bt: \(A=\dfrac{2008^{2008}+1}{2008^{2009}+1};B=\dfrac{2008^{2007}+1}{2008^{2008}+1}\)
\(A=\dfrac{2008^{2008}+1}{2008^{2009}+1}\)
\(2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}\)
\(=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}\)
\(=1+\dfrac{2007}{2008^{2009}+1}\)
\(B=\dfrac{2008^{2007}+1}{2008^{2008}+1}\)
\(2008\cdot B=\dfrac{2008^{2008}+2008}{2008^{2008}+1}\)
\(=\dfrac{2008^{2008}+1+2007}{2008^{2008}+1}\)
\(=1+\dfrac{2007}{2008^{2008}+1}\)
Ta có: \(2008^{2009}+1>2008^{2008}+1\)
\(\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}\)
\(\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}\)
\(\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}\)
hay \(A < B\)
#\(Toru\)
So sánh A và B biết \(A=\frac{2006}{2007}-\frac{2007}{2008}+\frac{2008}{2009}-\frac{2009}{2010};B=\frac{1}{2006.2007}-\frac{1}{2008.2009}\)
Tính tỉ số \(\dfrac{A}{B}\) , biết:
\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\)
\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)
\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)