tìm x biết |x-2007|-|x-2008|=1
Tìm x, biết :
X + x + 1 + x + 2 + ... + 2007 + 2008 = 2008
Tìm x, biết :
x + x + 1 + x + 2 + ... + 2007 + 2008 = 2008
(x+2007) + ( x+1+2006) + ..... +0 =0
=> x +2007 =0
=> x =-2007
Tìm XEZ biết
a)x+(x+1)+(x+2)+........+2008=2008
b)2009+2008+2007+........+(x+1)+x=2009
a)=> (2008+x).2008/2=2008
=>(2008+x)=2
=>x=-2006
Tìm x biết : x - 1 / 2009 + x - 2 / 2008 = x - 3 / 2007 + x- 4 / 2006
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)
\(\Leftrightarrow x=2010\)
\(\frac{\text{(2007−x)^2+(2007−x)(x−2008)+(x−2008)^2}}{\text{(2007−x)^2−(2007−x)(2008−x)+(x−2008)^2}}=\frac{19}{49}\)Tìm x
Ta có: \(\frac{\left(2007-x\right)^2+\left(2007-x\right)\left(x-2008\right)+\left(x-2008\right)^2}{\left(2007-x\right)^2-\left(2007-x\right)\left(2008-x\right)+\left(x-2008\right)^2}\)
\(=\frac{\left(2007-x\right)^2+\left(2007-x\right)\left(x-2008\right)+\left(x-2008\right)^2}{\left(2007-x\right)^2+\left(2007-x\right)\left(x-2008\right)+\left(x-2008\right)^2}\)
\(=1\)
Tìm x biết:
(x - 1)/2009 + (x - 2)/2008 = (x - 3)/2007 + (x - 4)/2006
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)
\(\Rightarrow x-2010=0\Rightarrow x=2010\)
Tìm x∈Z biết
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2007}=\dfrac{x+2}{2009}+\dfrac{x+1}{2010}\)
Tìm x,biết
x+4/2007 + x+3/2008 = x+2/2009 + x+1/2010
(x-4)/2007 + (x-3)/2008)= (x-2)/2009 + (x-1)/2010
=[(x-4)/2007 -1]+[(x-3)/2008 -1]=[(x-2)/2009 -1]+[(x-1)/2010 -1]
=(x-2011)/2007+(x-2011)/2008=(x-2011)/...
=(x-2011)(1/2007+1/2008-1/2009-1/2010)...
suy ra x=2010
tìm x biết giá trị tuyệt đối của x - 2007 - giá trị tuyệt đối của x - 2008 =1
X=2008;2009;2010;2011;2012;2013;2014;2015;2016;2017;.............................................