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3 tháng 8 2015
ta có \(1+\frac{x+5}{1995}+1+\frac{x+4}{1996}+1+\frac{x+3}{1997}=1+\frac{x+1995}{5}+1+\frac{x+1996}{4}+1+\frac{x+1997}{3}\)
\(=\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}=\frac{x+2000}{5}+\frac{x+2000}{4}+\frac{x+2000}{3}\)
\(=\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\right)=\left(x+2000\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\) (1)
Xét \(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\ne\frac{1}{5}+\frac{1}{4}+\frac{1}{3}vàx+2000=x+2000\) (2)
từ \(\left(1\right)\Leftrightarrow x+2000=0\) ( để (1) là đúng )
\(\Rightarrow x=2000\)
N
0
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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
HP
14 tháng 2 2016
cộng 1 vào mỗi tỉ số,ta đc:
(x+5)/1995+1+(x+4)/1996+1+(x+3)/1997+1=(x+1995)/5+1+(x+1996)/4+1+(x+1997|/3+1
=>\(\frac{x+5+1995}{1995}+\frac{x+4+1996}{1996}+\frac{x+3+1997}{1997}=\frac{x+1995+5}{5}+\frac{x+1996+4}{4}+\frac{x+1997+3}{3}\)
\(\Rightarrow\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}-\frac{x+2000}{5}-\frac{x+2000}{4}-\frac{x-2000}{3}=0\)
\(\Rightarrow\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
mà bt trong ngoặc thứ 2 khác 0
=>x+2000=0
=>x=-2000
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
23 tháng 7 2017
a. \(\frac{x-5}{2000}+\frac{x-4}{1999}+\frac{x-3}{1998}=\frac{x-2}{1997}+\frac{x-1}{1996}+\frac{x}{1995}\)
\(\Leftrightarrow\left(\frac{x-5}{2000}+1\right)+\left(\frac{x-4}{1999}+1\right)+\left(\frac{x-3}{1998}+1\right)=\left(\frac{x-2}{1997}+1\right)+\left(\frac{x-1}{1996}+1\right)+\left(\frac{x}{1995}+1\right)\)
\(\Leftrightarrow\left(x+1995\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}-\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)=0\)
\(\Leftrightarrow x+1995=0\)
\(\Leftrightarrow x=-1995\)
Ta có: \(\frac{x+3}{1995}+\frac{x+1}{1997}=-2\)
\(\Leftrightarrow\left(\frac{x+3}{1995}+1\right)+\left(\frac{x+1}{1997}+1\right)=0\)( cộng cả hai vế cho 2)
\(\Leftrightarrow\frac{x+1998}{1995}+\frac{x+1998}{1997}=0\)
\(\Leftrightarrow\left(x+1998\right).\left(\frac{1}{1995}+\frac{1}{1997}\right)=0\)
Mà \(\frac{1}{1995}+\frac{1}{1997}\ne0\)
\(\Rightarrow x+1998=0\)
\(\Rightarrow x=-1998\)