LP
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HG
1
HY
13 tháng 9 2016
\(\frac{x-1}{2010}+...+\frac{x-2010}{1}=2010\\ \Leftrightarrow\left(\frac{x-1}{2010}-1\right)+\left(\frac{x-2}{2009}-1\right)+...+\left(\frac{x-2010}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2011}{2010}+\frac{x-2011}{2009}+...+\frac{x-2011}{1}=0\)
\(\Leftrightarrow\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2009}+...+1\right)=0\)
\(\Leftrightarrow x-2011=0\) (Vì 1/2010 +1/2009 + ... +1 khác 0 )
\(\Leftrightarrow x=2011\)
LP
1
18 tháng 2 2017
\(\frac{x-1}{2010}+\frac{x-2}{2009}+....+\frac{x-2010}{2}=2010\)
\(\Leftrightarrow\left(\frac{x-1}{2010}-1\right)+\left(\frac{x-2}{2009}-1\right)+...+\left(\frac{x-2010}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2011}{2010}+\frac{x-2011}{2009}+....+\frac{x-2011}{1}=0\)
\(\Leftrightarrow\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2009}+...+1\right)=0\)
\(\Rightarrow x-2011=0\Rightarrow x=2011\)
LV
1
NV
24 tháng 5 2017
\(\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
BN
16 tháng 7 2017
\(\Leftrightarrow\frac{x+1}{2009}+\frac{x+1}{2010}+\frac{x+1}{2011}-\frac{x+1}{2012}-\frac{x+1}{2013}-\frac{x+1}{2014}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}=0\end{cases}}\)
mà \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
nên \(x+1=0\)
\(\Leftrightarrow x=-1\)
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.
2 tháng 4 2018
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+\frac{x-4}{2009}+\frac{x-2021}{2}=0\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+\left(\frac{x-4}{2009}-1\right)+\left(\frac{x-2021}{2}+4\right)=0\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+\frac{x-2013}{2009}+\frac{x-2013}{2}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2}\ne0\)
Nên \(x-2013=0\)
\(\Rightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt ~