LM
1
K
Khách
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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
NL
2
23 tháng 6 2016
http://olm.vn/hoi-dap/question/425074.html
dzô đó là có cách giải
LV
1
NV
24 tháng 5 2017
\(\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
TN
19 tháng 12 2016
\(\frac{x+1}{2013}+\frac{x}{2012}+\frac{x-1}{2011}=\frac{x-2}{2010}+\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Leftrightarrow\frac{x+1}{2013}-1+\frac{x}{2012}-1+\frac{x-1}{2011}-1=\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}=\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}-\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2012=0\). Do \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
\(\Leftrightarrow x=2012\)
26 tháng 9 2017
Cho x,y là các số nguyên dương, chứng minh rằng:
\(1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)
WJ
1
NQ
30 tháng 7 2015
\(\left(\frac{x+4}{2007}+1\right)+\left(\frac{x+3}{2008}+1\right)=\left(\frac{x+2}{2009}+1\right)+\left(\frac{x+1}{2010}+1\right)\)
\(\left(\frac{x+2011}{2007}\right)+\left(\frac{x+2011}{2008}\right)=\left(\frac{x+2011}{2009}\right)+\left(\frac{x+2011}{2010}\right)\)
\(\frac{x+2011}{2007}+\frac{x+2011}{2008}-\frac{x+2011}{2009}-\frac{x+2011}{2010}=0\)
\(\left(x+2011\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\)khác 0 (các số hạng ko bằng nhau)
\(\Leftrightarrow\)\(x+2011=0\)
\(\Rightarrow x=0-2011\)
\(\Rightarrow x=-2011\)
TY
8 tháng 8 2019
\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)
\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)
\(\Rightarrow|x-\frac{1}{3}|=2.8\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)
tính lại kết quả nhé
\(\frac{2-x}{2008}-1=\frac{1-x}{2009}-\frac{x}{2010}\)
\(\frac{2-x}{2008}-1+2=\frac{1-x}{2009}+1-\frac{x}{2010}+1\)
\(\frac{2-x}{2008}+1=\frac{1-x}{2009}+1+\frac{-x}{2010}+1\)
\(\frac{2-x}{2008}+\frac{2008}{2008}=\frac{1-x}{2009}+\frac{2009}{2009}+\frac{-x}{2010}+\frac{2010}{2010}\)
\(\frac{2010-x}{2008}=\frac{2010-x}{2009}+\frac{2010-x}{2010}\)
\(\frac{2010-x}{2008}-\frac{2010-x}{2009}-\frac{2010-x}{2010}=0\)
\(\left(2010-x\right).\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
\(\Rightarrow2010-x=0\text{ }\left(\text{vì }\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\ne0\right)\)
\(x=2010\)