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TN
5 tháng 3 2018
Ta có: x=2011 \(\Rightarrow\)x+1=2012
\(\Rightarrow A=x^{2011}-\left(x+1\right).x^{2010}\)\(+\left(x+1\right)x^{2009}\)\(-\left(x+1\right)x^{2008}+...\)\(-\left(x+1\right)x^2+\left(x+1\right)x-1\)
=\(x^{2011}\)\(-x^{2011}-x^{2010}+x^{2010}+x^{2009}-x^{2009}-\)...\(-x^2+x^2+x-1\)
= \(x-1=2011-1=2010\)
=
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. 
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 \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
\(\Rightarrow x+2014=0\)
\(\Rightarrow x=-2014\)
NH
0
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
HT
2
11 tháng 5 2015
Bỏ dấu giá trị tuyệt đối:
| x \(\le\) 2008 | 2008 < x < 2009 | 2009 \(\le\) x < 2010 | 2010\(\le\)x < 2011 | x \(\ge\) 2011 | |
| |x- 2008| | 2008-x | x-2008 | x-2008 | x-2008 | x-2008 |
| |x-2009| | 2009-x | 2009-x | x-2009 | x-2009 | x-2009 |
| |x-2010| | 2010-x | 2010 - x | 2010 - x | x - 2010 | x - 2010 |
| |x-2011| | 2011 - x | 2011 - x | 2011 - x | 2011 - x | x - 2001 |
=>
+) Nếu x \(\le\) 2008 => A = 2008 - x + 2009 - x + 2010 - x + 2011 - x + 2008 = 10 046 - 4x \(\ge\) 10 046 - 4.2008 = 2014
+) Nếu 2008 < x < 2009 => A = x - 2008 + 2009 - x + 2010 - x + 2011 - x + 2008 = 6030 - 2x > 6030 - 2.2009 = 2012
+) Nếu 2009 \(\le\) x < 2010 => A = x - 2008 + x - 2009 + 2010 - x + 2011 - x + 2008 = 2012
+) Nếu 2010 \(\le\) x < 2011 => A = x - 2008 + x - 2009 + x - 2010 + 2011 - x + 2008 = 2x - 2008 \(\ge\) 2.2010 - 2008 = 2012
+) Nếu x \(\ge\) 2011 => A = x - 2008 + x - 2009 + x - 2010 + x - 2011 + 2008 = 4x - 6030 \(\ge\) 4.2011 - 6030 = 2014
Từ các trường hợp trên => A nhỏ nhất bằng 2012 khi x = 2009 ; hoặc x = 2010
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
LH
0
VT
2
KN
28 tháng 10 2020
\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
\(\Leftrightarrow\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)
\(\Leftrightarrow\frac{x+2012}{2008}+\frac{x+2012}{2009}-\frac{x+2012}{2010}-\frac{x+2012}{2011}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2012=0\)
\(\Leftrightarrow x=-2012\)