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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
BN
19 tháng 2 2017
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2010}+1\right)-\left(\frac{x-3}{2009}+1\right)=\frac{x-4}{2008}+1\)
\(\Rightarrow\frac{x-1+2011}{2011}+\frac{x-2+2010}{2010}-\frac{x-3+2009}{2009}=\frac{x-4+2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x - 2012 = 0
=> x = 2012
Vậy x = 2012
30 tháng 1 2019
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2010}+1\right)-\left(\frac{x-3}{2009}+1\right)=\frac{x-4}{2008}+1\)
\(\Rightarrow\frac{x-1+2011}{2011}+\frac{x-2+2010}{2010}-\frac{x-3+2009}{2009}=\frac{x-4+2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x - 2012 = 0
=> x = 2012
Vậy x = 2012
31 tháng 1 2019
Kết quả đúng òi nhưng mà dấu suy ra thứ 2 ế \(x-1+2011\) thì bằng \(x+2010\) mà. Cả mấy cái bên cạnh cũng bị tính sai. 
QT
1
T
17 tháng 9 2018
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
LT
14 tháng 3 2017
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(=>\dfrac{x+4}{2010}+1\))+(\(\dfrac{x+3}{2011}+1\))=\(\left(\dfrac{x+2}{2012}+1\right)\)+\(\left(\dfrac{x+1}{2013}+1\right)\)
=>\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
=>x+2014(\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\))=0
ta thấy \(\dfrac{1}{2010}>\dfrac{1}{2011}>\dfrac{1}{2012}>\dfrac{1}{2013}\)
=>\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}>0\)
để A=0
\(\Leftrightarrow x+2014=0\)
\(\Leftrightarrow\)x=-2014
14 tháng 3 2017
a)\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)Mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
\(\Rightarrow x+2014=0\)
\(\Rightarrow x=-2014\)
TN
3
7 tháng 9 2015
(x + 4)/2010 + (x+3)/2011 = (x+2)/2012 + (x+1)/2013
<=> [(x + 4)/2010 + 1] + [(x+3)/2011 + 1] = [(x+2)/2012 + 1] + [(x+1)/2013 + 1]
<=> (x + 2014)/2010 + (x + 2014)/2011 = (x + 2014)/2012 + (x + 2014)/2013
<=> (x + 2014)/2010 + (x + 2014)/2011 - (x + 2014)/2012 - (x + 2014)/2013 = 0
<=> (x + 2014).(1/2010 + 1/2011 - 1/2012 - 1/2013) = 0
Ta thấy (1/2010 + 1/2011 - 1/2012 - 1/2013) ≠ 0
Vậy suy ra x = -2014
HD
4
HP
11 tháng 2 2016
trừ 1 vào mỗi tỉ số,ta đc:
\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}-1=\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}=\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(mà\frac{1}{2011}<\frac{1}{2010}<\frac{1}{2009}<\frac{1}{2008}\Rightarrow\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=>x-2012=0
=>x=2012
vậy x=2012