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\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)
\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)
Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức
\(a)\) \(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)
\(\Leftrightarrow\)\(\frac{x-1+2011}{2011}+\frac{x-2+2012}{2012}=\frac{x-3+2013}{2013}+\frac{x-4+2014}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}-\frac{x+2010}{2013}-\frac{x+2010}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow\)\(x=2010\)
Vậy \(x=2010\)
Chúc bạn học tốt ~
\(\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}\)\(=\frac{x-4}{2008}\)
\(\Leftrightarrow\frac{x-2012+2011}{2011}+\frac{x-2012+2010}{2010}+\frac{x-2012+2009}{2009}=\frac{x-2012+2008}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2011}+1+\frac{x-2012}{2010}+1+\frac{x-2012}{2009}+1=\frac{x-2012}{2008}+1\)
\(\Leftrightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}+2=\frac{x-2012}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2008}-\frac{x-2012}{2009}-\frac{x-2012}{2010}-\frac{x-2012}{2011}-2=0\)
=>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
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=0\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1=0\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)+\left(\frac{x-3}{2009}-1\right)+\left(\frac{x-4}{2008}-1\right)=0\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\cdot\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)\)
Vì \(\frac{1}{2011}< \frac{1}{2009}\) và \(\frac{1}{2010}< \frac{1}{2008}\) nên \(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\ne0\)
\(\rightarrow x-2012=0\)
\(\rightarrow x=2012\)
Vậy x = 2012.
http://olm.vn/hoi-dap/question/425074.html
dzô đó là có cách giải
\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)
\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)
\(\Rightarrow|x-\frac{1}{3}|=2.8\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)
tính lại kết quả nhé
\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
=> \(\left(\frac{x+4}{2008}+1\right)+\left(\frac{x+3}{2009}+1\right)=\left(\frac{x+2}{2010}+1\right)+\left(\frac{x+1}{2011}+1\right)\)
=> \(\frac{x+4+2008}{2008}+\frac{x+3+2009}{2009}=\frac{x+2+2010}{2010}+\frac{x+1+2011}{2011}\)
=> \(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)
=> \(\frac{x+2012}{2008}+\frac{x+2012}{2009}-\frac{x+2012}{2010}-\frac{x+2012}{2011}=0\)
=> \(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Do \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
=> x + 2012 = 0 => x = -2012
x+1/2008 + x+3/2009 = x+2/2010 + x+1/2011
suy ra : x+2012/2008 + x+2012/2009 = x+2012/2010 + x+2012/2011
Tương đương : (x+2012) . (1/2008 + 1/2009 = 1/2010 - 1/2011)=0
Tương đương : x+2012 = 0
suy ra : 0 - 2012
= -2012
Vậy x = -2012
Ta có : \(\frac{x+2011}{1}+\frac{x+2008}{2}+\frac{x+2007}{3}+\frac{x+2011}{5}=-15\)
\(\Rightarrow\left(\frac{x+2011}{1}+5\right)+\left(\frac{x+2008}{2}+4\right)+\left(\frac{x+2007}{3}+3\right)+\left(\frac{x+2008}{4}+2\right)+\left(\frac{x+2011}{5}+1\right)\)
\(=0\)
=> \(\frac{x+2016}{1}+\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{4}+\frac{x+2016}{5}=0\)
=> \(\left(x+2016\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=0\)
Vì \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\ne0\)
=> x + 2016 = 0
=> x = -2016
Vậy x = -2016
x+(x+2008)1/2+(x+2007)1/3+(x+2008)1/4+(x+2011)1/5=-15-2011=-2026
<=> x+x/2+1004+x/3+669+x/4+502+x/5+2011/5=-2026
<=>x+x/2+x/3+x/4+x/5+2011/5=-2026-1004-669-502=-4201
<=>x(1+(1)/(2)+(1)/(3)+(1)/(4)+(1)/(5))=-4201-(2011)/(5)=-23016/5
<=>x=-23016/5:(1+1/2+1/3+1/4+1/5)=-2016