TT
1
K
Khách
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NM
1
KT
15 tháng 8 2018
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
<=> \(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
<=> \(x+2004=0\) (do 1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0)
<=> \(x=-2004\)
HA
3
MH
5 tháng 10 2021
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)\)
⇔\(x+2014=0\)
⇔\(x=-2014\)
5 tháng 10 2021
\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\\ \Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\\ \Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\\ \Rightarrow x=-2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
LT
0
TT
0
11 tháng 9 2017
\(\dfrac{x+4}{2000}\) + \(\dfrac{x+3}{2001}\) =\(\dfrac{x+2}{2002}\) + \(\dfrac{x+1}{2003}\)
<=> \(\dfrac{x+4}{2000}\) + 1 + \(\dfrac{x+3}{2001}\) +1 = \(\dfrac{x+2}{2002}\) + 1 + \(\dfrac{x+1}{2003}\) + 1
<=>\(\dfrac{x+4}{2000}\)+\(\dfrac{2000}{2000}\)+\(\dfrac{x+3}{2001}\) \(\dfrac{2001}{2001}\) = \(\dfrac{x+2}{2002}\)+\(\dfrac{2002}{2002}\)+\(\dfrac{x+1}{2003}\)+\(\dfrac{2003}{2003}\)
<=> \(\dfrac{x+4+2000}{2000}\)+\(\dfrac{x+3+2001}{2001}\) = \(\dfrac{x+2+2002}{2002}\)+ \(\dfrac{x+1+2003}{2003}\)
<=> \(\dfrac{x+2004}{2000}\) + \(\dfrac{x+2004}{2001}\) - \(\dfrac{x+2004}{2002}\) - \(\dfrac{x+2004}{2003}\) = 0
<=> (x+2004)(\(\dfrac{1}{2000}\) + \(\dfrac{1}{2001}\) - \(\dfrac{1}{2002}\) -\(\dfrac{1}{2003}\)) = 0
mà \(\dfrac{1}{2000}\) + \(\dfrac{1}{2001}\) - \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\) khác 0
nên x+2004=0
=>x=0-2004
=> x = -2004
vậy S = -2004.
Tick nha
M
2
HP
25 tháng 9 2021
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}=0\)
<=> \(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)=0\)
<=> \(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
<=> \(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
<=> x + 2004 = 0
<=> x = -2004
(Bn nhớ thêm kết quả là 0 vào sau nữa nha)
CR
5
TT
13 tháng 9 2017
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\)\(\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)\)\(=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2004}{2000}\right)+\left(\frac{x+2004}{2001}\right)\)\(=\left(\frac{x+2004}{2002}\right)+\left(\frac{x+2004}{2003}\right)\)
\(\Leftrightarrow\left(x+2004\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)\(=0\)
\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=-2004\)
NV
1
X
22 tháng 1 2019
Sửa lại đề : Tìm x biết : \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Rightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2014}{2002}-\frac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left[\left(\frac{1}{2000}+\frac{1}{2001}\right)-\left(\frac{1}{2002}+\frac{1}{2003}\right)\right]=0\)
Mà : \(\frac{1}{2000}+\frac{1}{2001}>\frac{1}{2002}+\frac{1}{2003}\)
\(\Rightarrow x+2004=0\Rightarrow x=\left(-2004\right)\)
Ta có: \(\left(\frac{x+4}{2000}\right)+\left(\frac{x+3}{2001}\right)=\left(\frac{x+2}{2002}\right)+\left(\frac{x+1}{2003}\right)\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}\ne0\)
=> x + 2004 =0
=> x = -2004