1/(x+2000) - 1/(x+2007)=7/8
1/(x+2000) - 1/(x+2007)=7/8
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\(\Leftrightarrow\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{\left(x+2000\right)\left(x+2007\right)}\)
\(\Rightarrow\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{2^3}\)
\(\Rightarrow\frac{7\left(x^2+4007x+4013992\right)}{8\left(x+2000\right)\left(x+2007\right)}=0\)
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+2000)(x+2001) + 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) - 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
tìm x
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
Tìm các số nguyên x biết:
a,x+(x+1)+(x+2)+...+2006+2007=2007
b,2000+1999+...+x+1+x=2000
Lời gải:
a. Số số hạng:
$(2007-x):1+x=2008-x$
Suy ra:
$x+(x+1)+(x+2)+....+2006+2007=2007$
$\frac{(x+2007)(2008-x)}{2}=2007$
$(x+2007)(2008-x)=4014=$
$\Rightarrow x=2007$ hoặc $x=-2006$
b.
Số số hạng: $(2000-x):1+1=2001-x$
Suy ra:
$2000+1999+...+(x+1)+x=2000$
$\frac{(2000+x)(2001-x)}{2}=2000$
$(2000+x)(2001-x)=4000$
$\Rightarrow x=2000$ hoặc $x=-1999$
Tìm x biết :
a) x + ( x + 1 ) + ( x + 2) + ........+ ( x + 2006 ) +2007 =2007
b) 2000+ ( 199 +x ) + ( 198 + x ) + ( x + 1 ) + x = 200
a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 2006 ) + 2007 = 2007
\(\Rightarrow\)( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 2006 + 2007 ) = 2007
\(\Rightarrow\)2007x + 2015028 = 2007
\(\Rightarrow\)2007x = 2007 - 2015028 = -2013021
\(\Rightarrow\)x = ( -2013021 ) : 2007 = -1003
Vậy x = -1003
b) 2000 + ( 199 + x ) + ( 198 + x ) + ... + ( x + 1 ) + x = 200
\(\Rightarrow\)( x + x + x + ... + x + x ) + ( 1 + 2 + ... + 198 + 199 + 2000 ) = 200
\(\Rightarrow\)200x + 2001000 = 200
\(\Rightarrow\)200x = 200 - 2001000 = -2000800
\(\Rightarrow\)x = ( -2000800 ) : 200 = -10004
Vậy x = -10004
a, x + ( x + 1 ) + ( x + 2 ) + ..... + ( x + 2006) + 2007 = 2007
x. 2007 + ( 1 + 2 + ..... + 2006 ) = 2007 - 2007
x. 2007 + 2013021 = 0
x. 2007 = 0 - 2013021
x.2007 = - 2013021
x = ( - 2013021 ) : 2007
x = - 1003
a) x + ( x + 1 ) + ( x + 2) + ........+ ( x + 2006 ) +2007 =2007
(x+x+x+...+x)+(1+2+3+...+2007) =2007
2007.x + [(2007+1).2007 : 2] =2007
2007.x +2015028 =2007
2007.x =2007-2015028
x =(-2013021) : 2007
x = -1003
Vây x = -1003