tìm x biết
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
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OY
1
7 tháng 7 2017
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Rightarrow\left(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}\right)-\left(\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\right)=0\)
\(\Rightarrow\dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\)
\(\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\)
\(x+3=0\Rightarrow x=-3\)
7 tháng 7 2017
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\\ \dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\\ \left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x+3=0\Rightarrow x=-3\)
8 tháng 8 2018
\(\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}+\dfrac{x+5}{2007}+\dfrac{x+2007}{5}=-5\)
Ta có:
\(\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1+\dfrac{x+5}{2007}+1+\dfrac{x+2007}{5}+1=0\)
\(=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}+\dfrac{x+2012}{2007}+\dfrac{x+2012}{5}=0\)
\(=\left(x+2012\right)\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{5}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{5}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy \(x=-2012\)
Chúc bạn học tốt!
19 tháng 2 2022
\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)
=>x+2011=0
hay x=-2011
AH
Akai Haruma
Giáo viên
28 tháng 1 2021
Lời giải:
a)
PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)
\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)
\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)
\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)
\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)
b) Bạn kiểm tra lại xem có sai đề không?
GH
1
17 tháng 7 2022
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
TK
2
G
27 tháng 10 2017
Giải:
Ta có:
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\dfrac{x-1}{2009}+\dfrac{x-2}{2008}-2=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}-2\)
\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Leftrightarrow\dfrac{x-1-2009}{2009}+\dfrac{x-2-2008}{2008}=\dfrac{x-3-2007}{2007}+\dfrac{x-4-2006}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\Leftrightarrow\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow x=2010\)
Vậy \(x=2010\).
Chúc bạn học tốt!
27 tháng 10 2017
| \(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\) | ||||
| \(\Rightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\) | ||||
chuyển vế ta có:
|
NL
3
H
28 tháng 8 2017
Giải:
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Leftrightarrow\dfrac{x+4}{2008}+\dfrac{x+3}{2009}+2=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}+2\)
\(\Leftrightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Leftrightarrow\dfrac{x+4+2008}{2008}+\dfrac{x+3+2009}{2009}=\dfrac{x+2+2010}{2010}+\dfrac{x+1+2011}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên \(x+2012=0\)
\(\Leftrightarrow x=0-2012\)
\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\).
Chúc bạn học tốt!
28 tháng 8 2017
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Rightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên:
\(x+2012=0\Rightarrow x=-2012\)
\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Leftrightarrow x+3=0\)
hay x=-3