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Lời giải:
\(x=\frac{1}{2^{2009}}+\frac{2}{2^{2008}}+\frac{3}{2^{2007}}+....+\frac{2008}{2^2}+\frac{2009}{2}\)
\(2x = \frac{1}{2^{2008}}+\frac{2}{2^{2007}}+\frac{3}{2^{2006}}+...+\frac{2008}{2}+2009\)
\(\Rightarrow x=2x-x=2009-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2008}}-\frac{1}{2^{2009}}\)
\(\Rightarrow 2009-x=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\)
\(\Rightarrow 2(2009-x)=1+\frac{1}{2}+....+\frac{1}{2^{2007}}+\frac{1}{2^{2008}}\)
\(\Rightarrow 2(2009-x)-(2009-x)=1-\frac{1}{2^{2009}}\)
\(\Rightarrow 2009-x=1-\frac{1}{2^{2009}}\\ \Rightarrow x=2009-(1-\frac{1}{2^{2009}})=2008+\frac{1}{2^{2009}}\)
b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)
\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)
=> x + 2010 = 0 => x = -2010
Đặt cái trên là A nha
Ta có \(\left|A\right|=\left|-A\right|\ge A\)
nên |x-2005|+|x-2006|=|x-2005|+|2008-x| ≥ |x-2005+2008-x| ≥ |3|=3 (1)
mà |x-2005|+|x-2006|+|y-2007|+|x-2008|=3 (2)
từ (1) và (2) =>|x-2006|+|y-2007| ≤ 0 (*)
Để (*) xảy ra khi và chỉ khi x − 2006 = 0⇔x = 2006
y − 2007 = 0⇔y = 2007
( / x- 2005/ + / x-2008/ ) + / x-2006/ + / y +2007/ =3
VT =( / x -2005/ + / 2008 -x/ ) + / x -2006/ + / y+2007/ \(\ge\) / x-2005 + 2008 -x / + 0 + 0 = 3 = VP
Dấu ' =' xảy ra khi x -2006 =0 => x =2006
y + 2007 =0 => y =-2007
Vậy x =2006
y =-2007
(x+2007) + ( x+1+2006) + ..... +0 =0
=> x +2007 =0
=> x =-2007