2 So sanh
37/34... 27/24 2007/2005.... 2003/2001 203/201....423/601
K
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TD
0
10 tháng 6 2016
sao hai biểu thức đều có tên là E thế. biểu thức hai đặt tên lại là F nhé :
xét : E - F = \(\left(2000^2+2003^2+2005^2+2006^2\right)-\left(2001^2+2002^2+2004^2+2007^2\right).\)
\(=\left(2000^2-2001^2\right)+\left(2003^2-2002^2\right)+\left(2005^2-2004^2\right)+\left(2006^2-2007^2\right).\)
\(=-4001+4005+4009-4013=0\)
Vậy E = F
VT
2
18 tháng 4 2017
\(\dfrac{x+30}{2007}+\dfrac{x+32}{2005}=\dfrac{x+34}{2003}+\dfrac{x+36}{2001}\)
\(\Leftrightarrow\dfrac{x+30}{2007}+1+\dfrac{x+32}{2005}+1=\dfrac{x+34}{2003}+1+\dfrac{x+36}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2037}{2007}+\dfrac{x+2037}{2005}=\dfrac{x+2037}{2003}+\dfrac{x+2037}{2001}\)
\(\Leftrightarrow\dfrac{x+2037}{2007}+\dfrac{x+2037}{2005}-\dfrac{x+2037}{2003}-\dfrac{x+2037}{2001}=0\)
\(\Leftrightarrow\left(x+2037\right)\left(\dfrac{1}{2007}+\dfrac{1}{2005}-\dfrac{1}{2003}-\dfrac{1}{2001}\right)=0\)
\(\Rightarrow x+2037=0\).Do \(\dfrac{1}{2007}+\dfrac{1}{2005}-\dfrac{1}{2003}-\dfrac{1}{2001}\ne0\)
\(\Rightarrow x=-2037\)
VT
19 tháng 4 2017
Các bạn xem mình làm thế này có đúng không nhé. Nếu sai thì xin các bạn chữa hộ mình
Bài làm
\(\dfrac{x+30}{2007}+\dfrac{x+32}{2005}=\dfrac{x+34}{2003}+\dfrac{x+36}{2001}\)
\(\dfrac{x+30}{2007}+\dfrac{x+32}{2005}-\dfrac{x+34}{2003}-\dfrac{x+36}{2001}=0\)
\(\left(\dfrac{x+30}{2007}+1\right)+\left(\dfrac{x+32}{2005}+1\right)-\left(\dfrac{x+34}{2003}+1\right)-\left(\dfrac{x+36}{2001}+1\right)=0\)
\(\dfrac{x+30+2007}{2007}+\dfrac{x+32+2005}{2005}-\dfrac{x+34+2003}{2003}-\dfrac{x+36+2001}{2001}=0\)\(\dfrac{x+2037}{2007}+\dfrac{x+2037}{2005}-\dfrac{x+2037}{2003}-\dfrac{x+2037}{2001}=0\)\(\left(x+2037\right).\left(\dfrac{1}{2007}+\dfrac{1}{2005}-\dfrac{1}{2003}-\dfrac{1}{2001}\right)=0\)
x+2037=0
x = -2037
15 tháng 5 2017
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
\(A>8\)
16 tháng 8 2015
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
8 tháng 5 2018
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
27 tháng 2 2022
a, Ta có: \(\dfrac{27}{37}< \dfrac{27}{18};\dfrac{27}{18}< \dfrac{28}{18}\Rightarrow\dfrac{27}{37}< \dfrac{28}{18}\)
b, Ta có: \(1-\dfrac{2003}{2005}=\dfrac{2}{2005}\)
\(1-\dfrac{2001}{2003}=\dfrac{2}{2003}\)
Vì \(\dfrac{2}{2005}< \dfrac{2}{2003}\Rightarrow\dfrac{2003}{2005}>\dfrac{2001}{2003}\)
TC
4
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
\(\dfrac{1}{2001\times2003}+\dfrac{1}{2003\times2005}+...+\dfrac{1}{2011\times2013}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2001\times2003}+\dfrac{2}{2003\times2005}+...+\dfrac{2}{2011\times2013}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{1342671}\)
\(=\dfrac{2}{1342671}\)
T
16 tháng 10 2023
\(\dfrac{1}{2001\times2003}+\dfrac{1}{2003\times2005}+\dfrac{1}{2005\times2007}+...+\dfrac{1}{2011\times2013}\) (sửa đề)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{2001\times2003}+\dfrac{2}{2003\times2005}+\dfrac{2}{2005\times2007}+...+\dfrac{2}{2011\times2013}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+\dfrac{1}{2005}-\dfrac{1}{2007}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\times\dfrac{4}{1342671}\)
\(=\dfrac{2}{1342671}\)