\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)

\(\Leftrightarrow\frac{x+4}{2008}+1+\frac{x+3}{2009}+1-\frac{x+2}{2010}+1-\frac{x+1}{2011}+1=0\)

\(\Leftrightarrow\frac{x+4+2008}{2008}+\frac{x+3+2009}{2009}-\frac{x+2+2010}{2010}-\frac{x+1+2011}{2011}=0\)

\(\Leftrightarrow\frac{x+2012}{2008}+\frac{x+2012}{2009}-\frac{x+2012}{2010}-\frac{x+2012}{2011}=0\)

\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\right)=0\)

\(x=-2012\)

\(|y-3|\)= \(\frac{1}{\sqrt{4}}\) \(\Rightarrow\) \(|y-3|\)= \(\frac{1}{2}\)

\(\Rightarrow\)y - 3 = \(\frac{1}{2}\)hoặc y - 3 = \(\frac{-1}{2}\)

         y = \(\frac{1}{2}\)+ 3          y = \(\frac{-1}{2}\)+ 3

         y = \(\frac{7}{2}\)               y = \(\frac{5}{2}\)

Vậy có 2 giá trị của x là x = \(\frac{7}{2}\)hoặc x = \(\frac{5}{2}\).

Nhớ k cho mình nha! =)

|y-3|=1/2

y-3=1/2 hoặc y-3=-1/2

y=7/2 hoặc y= 5/2

Ta có \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{500}}\)

=> 5A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{499}}\)

Khi đó 5A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{499}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{500}}\right)\)

=> 4A = \(1-\frac{1}{5^{500}}\)

=> \(A=\frac{1}{4}-\frac{1}{4.5^{500}}< \frac{1}{4}\)

=> A < 1/4 (đpcm)

A=1/5+1/5^2+1/5^3+1/5^4+...+1/5^500

5A=1+1/5+1/5^2+1/5^3+...+1/5^499

5A-A=(1+1/5+1/5^2+1/5^3+...+1/5^499)-(1/5+1/5^2+1/5^3+1/5^4+...+1/5^500)

4A=1-1/5^500

A=(1-1/5^500)/4<1/4

A<1/4