Tìm x biết :\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\frac{1}{\left(x+2003\right)\left(x+2014\right)}=\frac{14}{15}\)
K
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26 tháng 5 2016
(x+4/2000 + 1)+(x+3/2001 + 1) = (x+2/2002 + 1)+(x+1/2003)+1
(x+2004/2000) + (x+2004/2001) = (x+2004/2002) + (x+2004/2003)
(x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
+ Với x+2004=0 suy ra x=-2004. Ta có 0.(1/2000+1/2001)=0.(1/2002+1/2003), đúng
+ Với x+2004 khác 0 thì (x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
1/2000+1/2001 = 1/2002+1/2003, vô lí vì 1/2000+1/2001 > 1/2002+1/2003
Vậy x=-2004
26 tháng 5 2016
\(PT\Leftrightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
<=> x + 2004 = 0
<=> x = -2004.
26 tháng 5 2016
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)
\(\Rightarrow x=-2004\)
13 tháng 7 2018
\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)
\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)
\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)
\(7\left(8-x^2-4007x-2000.2007\right)=0\)
\(8-x^2-4007x-2000.2007=0\)
\(x^2+4007x+4013992=0\)
\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)
\(\left(x+2008\right)\left(x+1999\right)=0\)
\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)
13 tháng 7 2018
\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
NM
31 tháng 1 2016
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)
=> x = -1999 hoặc x = - 2008
27 tháng 2 2019
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
MT
3 tháng 8 2015
\(\text{ĐKXĐ: }x+1\ne0\text{ và }x-2001\ne0\)
\(\Leftrightarrow x\ne-1\text{ và }x\ne2001\)
\(\frac{\left(x^2-2000x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x^2+x-2001x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}\)
\(=\frac{\left[x.\left(x+1\right)-2001\left(x+1\right)\right].2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x-2001\right)\left(x+1\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{2001}{2002}\)
Áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rút gọn rồi quy đồng làm nốt