\(3.\)

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

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

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

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

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

\(\Rightarrow\)\(x=2012\)

khó vậy

\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)  

\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)

\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)

\(\Rightarrow|x-\frac{1}{3}|=2.8\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)

tính lại kết quả nhé

\(\left|x-\frac{1}{3}\right|=\left|-3.2+\frac{2}{5}\right|=\left|-\frac{14}{5}\right|\)\(=\frac{14}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{14}{5}\\x-\frac{1}{3}=-\frac{14}{5}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{47}{15}\\x=-\frac{37}{15}\end{cases}}\)

Vậy............

b,

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

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

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

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

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

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> x-2012=0

=>x=2012

Vậy ..............

TK MK NHA

*****CHUC BN HOC GIỎI*****

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

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

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

\(\Leftrightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}+2=\frac{x-2012}{2008}\)

\(\Leftrightarrow\frac{x-2012}{2008}-\frac{x-2012}{2009}-\frac{x-2012}{2010}-\frac{x-2012}{2011}-2=0\)

=>Sai đề nha bạn!

áp dụng tính chất dãy tỷ số= nhau, ta có:

x-1/2011+x-2/2010+x-3/2009+x-4/2008=x-1+x-2+x-3+x-4/2011+2010+2009+2008

=x-1+x-2+x-3+x-4/8038

=(x-x+x-x)+[(1+4)+(-2+-3)]/8038

=0/8038

=0

a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)

=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)

=2.5

=10

a, Ta có \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)

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

<=> \(\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)-\left(\frac{x-3}{2009}-1\right)-\left(\frac{x-4}{2008}-1\right)=0\)

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

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

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> \(x-2012=0=>x=2012\)

b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)

=>\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}=2\cdot\frac{49}{99}\)

=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)

=>\(1-\frac{1}{2x+1}=\frac{98}{99}\)

=>\(\frac{2x}{2x+1}=\frac{98}{99}\)

=>2x = 98

=>x = 49

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

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

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

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

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

\(\Rightarrow x-2012=0\)

\(\Rightarrow x=2012\)

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)

\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)

\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)

\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)

x=-2011

cách bạn giải bài này là gì vậy ?

http://olm.vn/hoi-dap/question/425074.html

dzô đó là có cách giải

https://www.youtube.com/watch?v=LdnMg61kPHA