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a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
\(2012.\left|x-2011\right|+\left(x-2011\right)^2=2013\left|2011-x\right|\)
\(2012.\left|x-2011\right|+\left|x-2011\right|^2=2013\left|x-2011\right|\)
\(\left|x-2011\right|\left(2012+\left|x-2011\right|\right)=2013\left|x-2011\right|\)
\(\Rightarrow2012+\left|x-2011\right|=2013\)
\(\left|x-2011\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x=2012\\x=-2010\end{cases}}\)
Ta cÓ : ( x - 2011) x+ 1 - ( x - 2011)x + 2011
=) x - 2011= 0 =) x = 2011
\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)
\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)
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 ~
\(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2013}+1\right)+\left(\frac{x+2}{2012}+1\right)=\left(\frac{x+3}{2011}+1\right)+\left(\frac{x+4}{2010}+1\right)\)
\(\Rightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}-\frac{x+2014}{2011}-\frac{x+2014}{2010}=0\)
\(\Rightarrow\left(x+2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)nên để \(\left(x+2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Thì x+2014=0
=>x=-2014
\(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
=> \(\frac{x+1+2013}{2013}+\frac{x+2+2012}{2012}=\frac{x+3+2011}{2011}+\frac{x+4+2010}{2010}\)
=> \(\frac{x+2014}{2013}+\frac{x+2014}{2012}=\frac{x+2014}{2011}+\frac{x+2014}{2010}\)
=> \(\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
=> \(x+2014=0\)(do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\))
=> \(x=-2014\)
\(\Leftrightarrow2011.\left|x-2011\right|+\left(x-2011\right)^2=2013.\left|x-2011\right|\)
\(\Leftrightarrow\left(x-2011\right)^2=2.\left|x-2011\right|\Rightarrow\orbr{\begin{cases}x-2011=0\\2=\left|x-2011\right|\end{cases}}\Rightarrow\orbr{\begin{cases}x=2013\\x=2009\end{cases}}\text{hoặc }x=2011\)
Vậy ....