TN
3
K
Khách
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19 tháng 3 2020
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)
PK
0
VL
0
S
20 tháng 8 2017
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
=>\(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1-\frac{x-3}{2002}-1=\frac{x-4}{2001}-1\)
=> \(\frac{x-1-2004}{2004}+\frac{x-2-2003}{2003}-\frac{x-3-2002}{2002}=\frac{x-4-2001}{2001}\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}=\frac{x-2005}{2001}\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
=> \(\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Do \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
=> x - 2005 = 0 => x = 2005
NT
23
16 tháng 10 2016
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
\(\Rightarrow\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)
\(\Rightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1-\frac{x-3}{2002}+1-\frac{x-4}{2001}+1=0\)
\(\Rightarrow\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)
\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
\(\Rightarrow\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2004}< \frac{1}{2002};\frac{1}{2003}< \frac{1}{2001}\)\(\Rightarrow\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
\(\Rightarrow x-2005=0\)
\(\Rightarrow x=2005\)
Vậy x = 2005
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 \(x=2005\)
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
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
F=12.501.(x-4)
=4.501.3.(x-4)
=2004.\(\frac{2003}{2004}\)
=2003
\(F=12\left(501x-2004\right)\)
\(F=12.501\left(x-4\right)=6012.\frac{2003}{2004}=3.2003=6009\)