Giai phương trình
x+1/2009 + x+3/2007 = x+5/2005 + x+7/2003
K
Khách
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Những câu hỏi liên quan
5 tháng 3 2021
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
AH
Akai Haruma
Giáo viên
5 tháng 3 2021
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
C0
2
1 tháng 4 2018
Ta có :
\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)
\(\Leftrightarrow\)\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)
\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)
\(\Leftrightarrow\)\(\left(x-2014\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)
Nên \(x-2014=0\)
\(\Rightarrow\)\(x=2014\)
Vậy \(x=2014\)
Chúc bạn học tốt ~
1 tháng 4 2018
\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)
Trừ cả 2 vế cho 2 ta được :
\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)
\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)
\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)
\(\Leftrightarrow\left(x-2014\right)\times\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)
Mà : \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
11 tháng 4 2020
Đúng rồi bạn nhé! Đây là dạng toán quen thuộc nên có lẽ bạn trên viết nhầm đề nha!
S
6 tháng 2 2020
\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)'
\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+7}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{2003}\)
\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\left(vì:\frac{1}{2009}+\frac{1}{2007}< \frac{1}{2005}+\frac{1}{2003}\right)\)
CC
1
HN
26 tháng 1 2019
\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)
\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)
|________________A________________|
Do A > 0
nên x - 2014 = 0
<=> x = 2014
HD
1
17 tháng 6 2023
=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009
=2009
K
1
4 tháng 5 2019
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)
<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)
<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0
<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0
<=> x+2011=0
<=> x=-2011
Vậy pt có nghiệm là x=-2011
MM
Giải các phương trình:
\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)
1
27 tháng 1 2019
\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)
\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}-4=0\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-7}{2007}-1\right)+\left(\dfrac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2007}+\dfrac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\right)=0\)
\(\Leftrightarrow x-2014=0\) ( do \(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\ne0\))
\(\Leftrightarrow x=2014\)
Vậy phương trình có nghiệm S=\(\left\{2014\right\}\)
Ta có: \(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)
\(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+3}{2007}+1=\frac{x+5}{2005}+1+\frac{x+7}{2003}+1\)
\(\Leftrightarrow\frac{x+1+2009}{2009}+\frac{x+3+2007}{2007}=\frac{x+5+2005}{2005}+\frac{x+7+2003}{2003}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{2003}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}-\frac{x+2010}{2005}-\frac{x+2010}{2003}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)
=> x + 2010 = 0
=> x = -2010
Vậy x = -2010
\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)
\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+7}{2003}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2010}{2009}\right)+\left(\frac{x+2010}{2007}\right)=\left(\frac{x+2010}{2005}\right)+\left(\frac{x+2010}{2003}\right)\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\))
\(\Leftrightarrow x=-2010\)
Vậy tập nghiệm của phương trình là S = { -2010 } .