tìm x
1/(x+2001) - 1/(x+2007) =7/8
1/(x+2000)(x+2001) + 1/(x+2001)(x+2002) +1/(x+2002)(x+2003) +........+ 1/(x+2006)(x+2007)= 7/8
1/(x+2001) - 1/(x+2007) =7/8
\(\Leftrightarrow\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{\left(x+2000\right)\left(x+2007\right)}\)
\(\Rightarrow\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{2^3}\)
\(\Rightarrow\frac{7\left(x^2+4007x+4013992\right)}{8\left(x+2000\right)\left(x+2007\right)}=0\)
áp dụng Delta ta có :
\(\Leftrightarrow x^2+4007x+4013992=0\)
\(\Rightarrow4007^2-4\left(1.4013992\right)=81\)
\(\Rightarrow x_{1,2}=\frac{-b+-\sqrt{D}}{2a}=\frac{-4007+-\sqrt{81}}{2}\)
=>x=-2008 hoặc -1999
1/(x+2001)(x+2002)+1/(x+2002)(x+2003)+..........+1/(x+2006)(x+2007) =7/8
\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)
\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)
\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)
\(7\left(8-x^2-4007x-2000.2007\right)=0\)
\(8-x^2-4007x-2000.2007=0\)
\(x^2+4007x+4013992=0\)
\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)
\(\left(x+2008\right)\left(x+1999\right)=0\)
\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)
\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
phần đầu mk thiếu điều kiện,bn tự bổ sung nha
1 / giải phương trình sau:
\(\frac{1}{\left(x+2000\right).\left(x+2001\right)}+\frac{1}{\left(x+2001\right).\left(x+2002\right)}...\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)
=> x = -1999 hoặc x = - 2008
1/(x+2001)(x+2002) +1/(x+2002)(x+2003)+(1/(x+2003)(x+2004)+.......+ 1/(x+2006)(x+2007) =7/8
giải giúp mình chi tiết nha.
1-3+5-7+9-...-1999-2000+2001
1+2+3-4+5+6+7-8+...+1999-2000 + 2001
B= 1-7+13-19+25-31+....; biếtB có 2007 số hạng
1/2001 x 2003 +1 /2003 x 2005+ 1 /2005x 2007+...+1/ 2011 + 2013
\(\dfrac{1}{2001\times2003}+\dfrac{1}{2003\times2005}+...+\dfrac{1}{2011\times2013}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2001\times2003}+\dfrac{2}{2003\times2005}+...+\dfrac{2}{2011\times2013}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{1342671}\)
\(=\dfrac{2}{1342671}\)
\(\dfrac{1}{2001\times2003}+\dfrac{1}{2003\times2005}+\dfrac{1}{2005\times2007}+...+\dfrac{1}{2011\times2013}\) (sửa đề)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{2001\times2003}+\dfrac{2}{2003\times2005}+\dfrac{2}{2005\times2007}+...+\dfrac{2}{2011\times2013}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+\dfrac{1}{2005}-\dfrac{1}{2007}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\times\dfrac{4}{1342671}\)
\(=\dfrac{2}{1342671}\)
\(\dfrac{2009-x}{7}+\dfrac{2007-x}{9}+\dfrac{2005-x}{11}+\dfrac{2003-x}{13}=\dfrac{x-17}{-1999}+\dfrac{x-15}{-2001}+\dfrac{x-13}{-2003}+\dfrac{x-11}{-2005}\)