Tìm x biết
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
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EO
0
EO
1
PG
26 tháng 3 2017
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}\) + 35 = \(^{2^5}\)
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}\) = -3
\(\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)\) = 0
\(\left(\frac{x+1}{2004}+\frac{2004}{2004}\right)+\left(\frac{x+2}{2003}+\frac{2003}{2003}\right)+\left(\frac{x+3}{2002}+\frac{2002}{2002}\right)\)= 0
\(\left(\frac{x+2005}{2004}\right)+\left(\frac{x+2005}{2003}\right)+\left(\frac{x+2005}{2002}\right)\)= 0
\(\left(x+2005\right).\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)\) = 0
\(\left(x+2005\right)\) = 0 \(:\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)\)
\(\left(x+2005\right)\) = 0
\(x\) = 0-2005
\(x\) = -2005
CB
1
23 tháng 9 2016
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
=> \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)
=> \(\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
=> \(\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2004}< \frac{1}{2002}\); \(\frac{1}{2003}< \frac{1}{2001}\)
=> \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
=> \(x-2005=0\)
=> \(x=2005\)
Vậy \(x=2005\)
HJ
1
7 tháng 2 2018
Ta có: \(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)
\(\Leftrightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)
\(\Leftrightarrow\frac{x-1-2004}{2004}+\frac{x-2-2003}{2003}=\frac{x-3-2002}{2002}+\frac{x-4-2001}{2001}\)
\(\Leftrightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
=> x - 2005 = 0
=> x = 2005
Vậy x = 2005
H
2
1 tháng 2 2018
=> (x - 1)/2004 - 1 + (x - 2)/2003 - 1 = (x - 3)/2002 -1 + (x - 4)/2001 - 1
=> (x - 2005)/2004 + (x - 2005)/2003 = (x - 2005)/2002 + (x - 2005)/2001
=> (x - 2005)/2004 + (x - 2005)/2003 - (x - 2005)/2002 - (x - 2005)/2001 = 0
=> (x - 2005) * ( 1/2004 + 1/2003 - 1/2002 - 1/2001) = 0
Ta thấy ( 1/2004 + 1/2003 - 1/2002 - 1/2001) khác 0
=> x - 2005 = 0
=> x = 2005
D
1 tháng 2 2018
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
\(\Leftrightarrow\)\(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-4}{2001}+\frac{x-3}{2002}\)
\(\Leftrightarrow\)\(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\)\(\frac{x-4}{2001}-1+\frac{x-3}{2002}-1\)
\(\Leftrightarrow\)\(\frac{x-2005}{2004}+\frac{x-2005}{2003}\)\(=\frac{x-2015}{2001}+\frac{x-2005}{2002}\)
\(\Leftrightarrow\)\(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2001}-\frac{x-2005}{2002}=0\)
\(\Leftrightarrow\)( x - 2005 ) ( \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{2002}\)) = 0
Do \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{2002}\)\(\ne\)0
\(\Rightarrow\)x - 2005 = 0
\(\Leftrightarrow\)x = 2005
Vậy x = 2005
PN
14
19 tháng 11 2019
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
De thay \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}< 0\Rightarrow x+2005=0\)
\(\Rightarrow x=-2005\)
M
19 tháng 11 2019
Bài giải
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Do : \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
\(\Rightarrow\text{ }x+2005=0\)
\(x=0-2005\)
\(x=-2005\)
DH
0
NL
1
25 tháng 5 2016
\(\frac{\sqrt{x-2002}}{x-2002}-\frac{1}{x-2002}+\frac{\sqrt{y-2003}}{y-2003}-\frac{1}{y-2003}+\frac{\sqrt{z-2004}}{z-2004}-\frac{1}{z-2004}=\frac{3}{4}\)
\(1-\frac{1}{x-2002}+1-\frac{1}{y-2003}+1-\frac{1}{z-2004}=\frac{3}{4}\)
\(3-\frac{1}{x-2002}-\frac{1}{y-2003}-\frac{1}{z-2004}=\frac{3}{4}\)
\(\frac{1}{x-2002}+\frac{1}{y-2003}+\frac{1}{z-2004}=3-\frac{3}{4}=\frac{9}{4}\)
=> không có giá trị x,y,z thỏa mãn đề
Ta có :
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
\(\Leftrightarrow\)\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)=32-35+3\)
\(\Leftrightarrow\)\(\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=-3+3\)
\(\Leftrightarrow\)\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)
Nên \(x+2005=0\)
\(\Rightarrow\)\(x=-2005\)
Vậy \(x=-2005\)
Chúc bạn học tốt ~
Ta có: \(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)
\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=-3\)
\(\Rightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1=-3+3\)
\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}+\frac{x+3+2002}{2002}=0\)
\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=0\)
\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)
Nên x + 2005 = 0
=> x = -2005
Vậy x = -2005