\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)
\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)
\(\Rightarrow|x-\frac{1}{3}|=2.8\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)
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*****
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)=\left(\frac{x-3}{2009}-1\right)+\left(\frac{x-4}{2008}-1\right)\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\cdot\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
\(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
\(\Rightarrow\frac{1}{4}A=\frac{1}{2^4}-\frac{1}{2^6}+\frac{1}{2^8}-\frac{1}{2^{10}}+....+\frac{1}{2^{2004}}-\frac{1}{2^{2006}}\)
\(\Rightarrow\frac{5}{4}A=\frac{1}{2^2}-\frac{1}{2^{2006}}\)
\(\Rightarrow\frac{5}{4}A< \frac{1}{4}\Rightarrow5A< 1\Rightarrow A< \frac{1}{5}=0,2\)
P/S:Ai àm so ry bi cọt "hơi chậm":V