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\(A=\dfrac{1}{\left(-1997\right)\left(-1995\right)}+...+\dfrac{1}{\left(-3\right)\left(-1\right)}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{1995.1997}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{1995.1997}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{1995}-\dfrac{1}{1997}\right)\)
\(=\dfrac{1}{2}.\dfrac{1996}{1997}=\dfrac{998}{1997}\)
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
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
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 )
\(\Rightarrow x=2000\)
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 \(\frac{1}{1995}+\frac{1}{1997}\ne0\)
\(\Rightarrow x+1998=0\)
\(\Rightarrow x=-1998\)