Tìm x : x+5/2005 + x+6/2004 + x+7/2003 = -3
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DA
1
H
16 tháng 7 2023
\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\\ \Rightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\\ \Rightarrow x=-2010\)
NT
5
TH
30 tháng 6 2015
\(\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)
<=> \(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
<=>\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
<=>x+2010=0
<=>x=-2010
6 tháng 9 2017
\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\)
=>\(\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\)
=>\(\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)
=>\(x+2010=0\)(do\(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\)khác 0)
=>x=-2010
Vậy...
TN
1
8 tháng 8 2019
\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)
\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)
\(\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=0\)
\(\frac{x+5+2005}{2005}+\frac{x+6+2004}{2004}+\frac{x+7+2003}{2003}=0\)
\(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2012}{2003}=0\)
\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)
\(x+2010=0\)
\(x=-2010\)
D
23 tháng 5 2017
Bài 1:
Ta có: \(\frac{497}{-499}=-\frac{497}{499}>-\frac{499}{499}=-1\left(1\right)\)
\(-\frac{2345}{2341}< -\frac{2341}{2341}=-1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{497}{-499}>-\frac{2345}{2341}\)
Bài 2:
\(\frac{x+5}{2005}+\frac{x+6}{2004}=\frac{x+7}{2003}+3=0\)
\(\Rightarrow\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)
\(\Rightarrow\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)
\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
\(\Rightarrow\left(x+2010\right)\times\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
Vì \(\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\ne0\Rightarrow x+2010=0\)
\(\Rightarrow x=0-2010=-2010\)
Vậy x = -2010
\(\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=-3+3=0\)
\(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
=> x + 2010 = 0
=> x = -2010