AN
0
K
Khách
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TT
1
18 tháng 9 2016
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
NV
4
HH
9 tháng 3 2015
A=(1+2-3)+(-4+5+6-7)+(-8+9+10-11)+......(-2000+2001+2002-2003)
A=0+0....+0
A=0
NB
9 tháng 3 2015
Ta thấy 2-3-4=-5
6-7-8=-9
.............
1998-1999-2000=-2001
=> 1+2-3-4+5+6-7-8+....-1999-2000+2001-2003=1-5+5-9+9-...-2001+2001+2002-2003
=> A= 1+2002-2003=0
Vậy A=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
DK
5
BV
0
LH
2
10 tháng 1 2018
D = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 1999 - 2000 + 2001 + 2002 - 2003
D = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ... + ( 1997 + 1998 - 1999 - 2000 ) + 2001 + 2002 - 2003
D = ( -4 ) + ( -4 ) + ... + ( -4 ) + ( 2001 + 2002 - 2003 )
D = ( -4 ) . 500 + 2000
D = -2000 + 2000
D = 0
10 tháng 1 2018
D = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ............. - 1999 - 2000 + 2001 + 2002 - 2003
D = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ............ + ( 1997 + 1998 - 1999 - 2000 ) + 2001 + 2002 - 2003
D = ( -4 ) + ( -4 ) + .............. + ( -4 ) + ( 2001 + 2002 - 2003 )
D = ( -4 ) . 500 + 2000
D = -2000 + 2000
D = 0
DD
2
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)
<=> \(x=-2004\)