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NH
1
YN
16 tháng 9 2021
\(\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}=6\)
\(\Rightarrow\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}-6=0\)
\(\Rightarrow\left(\frac{x-1}{2000}-1\right)+\left(\frac{x-3}{1998}+1\right)+\left(\frac{x-5}{1996}-1\right)+\left(\frac{x}{667}-3\right)=0\)
\(\Rightarrow\frac{x-1-2000}{2000}+\frac{x-3-1998}{1998}+\frac{x-5-1996}{1996}+\frac{x-3.667}{667}=0\)
\(\Rightarrow\frac{x-2001}{2000}+\frac{x-2001}{1998}+\frac{x-2001}{1996}+\frac{x-2001}{667}=0\)
\(\Rightarrow\left(x-2001\right)\left(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\right)=0\)
Ta có: \(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\ne0\)
\(\Rightarrow x-2001=0\Rightarrow x=2001\)
LT
25 tháng 2 2020
\(\left(\frac{x-10}{1994}-1\right)\)+\(\left(\frac{x-8}{1996}-1\right)\)+\(\left(\frac{x-6}{1998}-1\right)\)+\(\left(\frac{x-4}{2000}-1\right)\)+\(\left(\frac{x-2}{2002}-1\right)\)=\(\left(\frac{x-2002}{2}-1\right)\)+\(\left(\frac{x-2000}{4}-1\right)\)+\(\left(\frac{x-1998}{6}-1\right)\)+\(\left(\frac{x-1996}{8}-1\right)\)+\(\left(\frac{x-1994}{10}-1\right)\)
suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)=\(\frac{x-2004}{2}\)+\(\frac{x-2004}{4}\)+\(\frac{x-2004}{6}\)+\(\frac{x-2004}{8}\)+\(\frac{x-2004}{10}\)
suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)- \(\frac{x-2004}{2}\)- \(\frac{x-2004}{4}\)- \(\frac{x-2004}{6}\)- \(\frac{x-2004}{8}\)- \(\frac{x-2004}{10}\)=0
suy ra (x-2004) . ( \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\))=0
Vì \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\) khác 0
nên x-2004=0 suy ra x=2004
NB
1
30 tháng 3 2019
Suy ra \(\frac{x+1}{1999}+1+\frac{x+2}{1998}+1=\frac{x+3}{1997}+1+\frac{x+4}{1996}\)
Suy ra \(\frac{x+2000}{1999}+\frac{x+2000}{1998}=\frac{x+2000}{1997}+\frac{x+2000}{1996}\)
Suy ra \(\frac{x+2000}{1999}+\frac{x+2000}{1998}-\frac{x+2000}{1997}-\frac{x+2000}{1996}=0\)
Suy ra \(x+2000.\left(\frac{1}{1999}+\frac{1}{1998}-\frac{1}{1997}-\frac{1}{1996}\right)=0\)
Vì \(\left(\frac{1}{1999}+\frac{1}{1998}-\frac{1}{1997}-\frac{1}{1996}\right)\ne0\)
Suy ra x+2000=0
Suy ra x=-2000
Hok tốt
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 8 2023
\(\dfrac{x-6}{1998}\) + \(\dfrac{x-4}{2000}\) = \(\dfrac{x-2000}{4}\) + \(\dfrac{x-1998}{6}\)
\(\dfrac{x-6}{1998}\) - 1 + \(\dfrac{x-4}{2000}\) - 1 = \(\dfrac{x-2000}{4}\) - 1 + \(\dfrac{x-1998}{6}\) - 1
\(\dfrac{x-6-1998}{1998}\) + \(\dfrac{x-4-2000}{2000}\) = \(\dfrac{x-2000-4}{4}\) + \(\dfrac{x-1998-6}{6}\)
\(\dfrac{x-2004}{1998}\) + \(\dfrac{x-2004}{2000}\) = \(\dfrac{x-2004}{4}\) + \(\dfrac{x-2004}{6}\)
(\(x-2004\)).[\(\dfrac{1}{1998}\) + \(\dfrac{1}{2000}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)] = 0
\(x\) - 2004 = 0
\(x\) = 2004
2 tháng 10 2017
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
DT
6 tháng 4 2018
dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
KT
12 tháng 8 2018
1) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
<=> \(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
<=> \(x+1=0\) (do 1/2 + 1/3 + 1/4 - 1/5 - 1/6 khác 0)
<=> \(x=-1\)
Vậy...
KT
12 tháng 8 2018
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
<=> \(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> \(x+2010=0\) (do 1/2009 + 1/2008 + 1/2007 - 1/2000 - 1/1999 - 1/1998 khác 0)
<=> \(x=-2010\)
Vậy....
12 tháng 8 2018
1,
x+1/2+x+1/3+x+1/4-x+1/5-x+1/6=0
(x+1)(1/2+1/3+1/4-1/5-1/6)=0
vì 1/2+1/3+1/4-1/5-1/6 khác 0
suy ra x+1=0 suy ra x=-1