x-1/2011+x-2/2010-x-3/2009=x-4/2008
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NT
3
10 tháng 7 2016
(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (1 + 1/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (4/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x 0
=0
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. 
HP
28 tháng 9 2021
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}-\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
<=> \(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-3}{2009}-1\right)=\left(\dfrac{x-4}{2008}-1\right)\)
<=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
<=> x - 2012 = 0
<=> x = 2012
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
NT
1
1 tháng 4 2022
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
=>x-2014=0
hay x=2014
24 tháng 7 2017
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\)
\(\Rightarrow\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1=0\)
\(\Rightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy x = -2012
MV
24 tháng 7 2017
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\\ \Leftrightarrow1+\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}=0\\ \Leftrightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\\ \Leftrightarrow \left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\\ \Rightarrow x+2012=0\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}>0\right)\\ \Rightarrow x=-2012\)
Vậy \(x=-2012\)
VT
2
KN
28 tháng 10 2020
\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
\(\Leftrightarrow\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)
\(\Leftrightarrow\frac{x+2012}{2008}+\frac{x+2012}{2009}-\frac{x+2012}{2010}-\frac{x+2012}{2011}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2012=0\)
\(\Leftrightarrow x=-2012\)
LT
1
YN
2 tháng 3 2022
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
VT
2
TT
8 tháng 2 2020
\(\frac{x}{2008}+\frac{x+1}{2009}+...+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+...+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+...+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)=0\)
Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)\ne0\)
Nên \(x-2008=0\)
\(\Leftrightarrow x=2008\)
Vậy : \(x=2008\)
CC
8 tháng 2 2020
\(\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}-5=0\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+\left(\frac{x+2}{2010}-1\right)+\left(\frac{x+3}{2011}-1\right)+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+\frac{x-2008}{2010}+\frac{x-2008}{2011}+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\ne0\)
\(\Rightarrow x-2008=0\)\(\Leftrightarrow x=2008\)
Vậy \(x=2008\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2010}+1\right)-\left(\frac{x-3}{2009}+1\right)=\frac{x-4}{2008}+1\)
\(\Rightarrow\frac{x-1+2011}{2011}+\frac{x-2+2010}{2010}-\frac{x-3+2009}{2009}=\frac{x-4+2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x - 2012 = 0
=> x = 2012
Vậy x = 2012
phải là trừ 1 chứ sao lại là cộng 1