\(\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{2003}\)'
\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+7}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2007}=\frac{x+2010}{2005}+\frac{x+2010}{2003}\)
\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\left(vì:\frac{1}{2009}+\frac{1}{2007}< \frac{1}{2005}+\frac{1}{2003}\right)\)
Giai phương trình
x+1/2009 + x+3/2007 = x+5/2005 + x+7/2003
Giải các phương trình:
\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)
\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)
CMĐT: 1996 x 1989 -1993 x 1998 x 1995 = 2005 x 2010 x 2003 - 2007 x 2002 x2009.
Giải Phương trình sau:
a:(x+1)/4-(x+2)/5+(x+4)/7-(x+5)/8+(x+7)/10-(x+9)/12=0
b:x/2004+(x+1)/2005+(x+2)/2006+(x+3)/2007=4
giai cac phuong trinh sau :
a, 1/x^2+5x+4+1/x^2+11x+28+1/x^2+17x+70=3/6x-2005
b, x+1/x-1+x-2/x+2+x-3/x+3+x+4/x-4=4
c, 1/4x-2006+1/5x+2004=1/15x-2007-1.6x-2005
d, 4x+16/x+6-3/x+1=5/x+3+7/x+5
Chứng minh rằng:
1996×1991×1989-1993×1988×1995=2005×2010×2003-2007×2002×2009
\(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
x-1/2009+x-2/2008=x-3/2007+x-4/2006
