\(\frac{x-1}{2001}\)+\(\frac{x-2}{2010}\)-\(\frac{x-3}{2009}\)=\(\frac{x-4}{2008}\)
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28 tháng 4 2018
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
LV
1
NV
24 tháng 5 2017
\(\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
CW
1 tháng 2 2017
\(\frac{x-2009-2010}{2008}+\frac{x-2008-2010}{2009}+\frac{x-2008-2009}{2010}=3\)
\(\Leftrightarrow\frac{x-2008-2009-2010}{2008}+\frac{x-2008-2009-2010}{2009}+\frac{x-2008-2009-2010}{2010}=0\)
\(\Leftrightarrow\left(x-2008-2009-2010\right)\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
\(\Leftrightarrow x-6027=0\Leftrightarrow x=6027\)
21 tháng 3 2017
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
HQ
2
13 tháng 10 2018
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}\)\(=\frac{x-4}{2008}\)
\(\Leftrightarrow\frac{x-2012+2011}{2011}+\frac{x-2012+2010}{2010}+\frac{x-2012+2009}{2009}=\frac{x-2012+2008}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2011}+1+\frac{x-2012}{2010}+1+\frac{x-2012}{2009}+1=\frac{x-2012}{2008}+1\)
\(\Leftrightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}+2=\frac{x-2012}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2008}-\frac{x-2012}{2009}-\frac{x-2012}{2010}-\frac{x-2012}{2011}-2=0\)
=>Sai đề nha bạn!
1 tháng 1 2020
áp dụng tính chất dãy tỷ số= nhau, ta có:
x-1/2011+x-2/2010+x-3/2009+x-4/2008=x-1+x-2+x-3+x-4/2011+2010+2009+2008
=x-1+x-2+x-3+x-4/8038
=(x-x+x-x)+[(1+4)+(-2+-3)]/8038
=0/8038
=0
HM
2
T
26 tháng 3 2019
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=0\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1=0\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)+\left(\frac{x-3}{2009}-1\right)+\left(\frac{x-4}{2008}-1\right)=0\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\cdot\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)\)
Vì \(\frac{1}{2011}< \frac{1}{2009}\) và \(\frac{1}{2010}< \frac{1}{2008}\) nên \(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\ne0\)
\(\rightarrow x-2012=0\)
\(\rightarrow x=2012\)
Vậy x = 2012.
DL
25 tháng 6 2018
\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)
\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)
Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức
25 tháng 6 2018
\(a)\) \(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)
\(\Leftrightarrow\)\(\frac{x-1+2011}{2011}+\frac{x-2+2012}{2012}=\frac{x-3+2013}{2013}+\frac{x-4+2014}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}-\frac{x+2010}{2013}-\frac{x+2010}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow\)\(x=2010\)
Vậy \(x=2010\)
Chúc bạn học tốt ~
Lời giải:
$\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1$
$\Leftrightarrow \frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}$
$\Leftrightarrow (x-2012)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0$
Dễ thấy $\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}< 0$
Do đó $x-2012=0\Rightarrow x=2012$
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}.\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-4}{2008}+\frac{x-3}{2009}\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)=\left(\frac{x-4}{2008}-1\right)+\left(\frac{x-3}{2009}-1\right)\)
\(\Rightarrow\left(\frac{x-1-2011}{2011}\right)+\left(\frac{x-2-2010}{2010}\right)=\left(\frac{x-4-2008}{2008}\right)+\left(\frac{x-3-2009}{2009}\right)\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2008}+\frac{x-2012}{2009}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2008}-\frac{x-2012}{2009}=0\)
\(\Rightarrow\left(x-2012\right).\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\ne0.\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=0+2012\)
\(\Rightarrow x=2012\)
Vậy \(x=2012.\)
Chúc bạn học tốt!