\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)

\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)

\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)

\(\frac{x-2015}{2014}+\frac{x-2015}{2013}-\frac{x-2015}{2012}-\frac{x-2015}{2011}=0\)

\(\left(x-2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)

\(\Rightarrow x-2015=0\)

\(x=0+2015\)

\(x=2015\)

\(x=2015\)

x=2015 nha

\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)

\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)

\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)

\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)

mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)

Vậy x=-2010

Ta có : \(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+\dfrac{x+3}{2012}+\dfrac{x+4}{2011}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+4}{2011}+1\right)=4\)

\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}=4\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}\right)=4\) \(\Leftrightarrow\left(x+2015\right).0,002=4\) ( mik lấy gần bằng nha )

\(\Leftrightarrow x+2015=2000\Leftrightarrow x=-15\)

Vậy phương trình có nghiệm là x=-15

\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)

\(\Leftrightarrow\left(\dfrac{x}{2010}-1\right)+\left(\dfrac{x+1}{2011}-1\right)+\left(\dfrac{x+2}{2012}-1\right)+\left(\dfrac{x+3}{2013}-1\right)+\left(\dfrac{x+4}{2014}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}+\dfrac{x-2010}{2012}+\dfrac{x-2010}{2013}+\dfrac{x-2010}{2014}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)=0\)

\(\Leftrightarrow x=2010\)

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

`=> (x+2014) (1/2010 + 1/2011-1/2012-1/2013)=0`

`=> x+2014=0` ( vì `1/2010 + 1/2011-1/2012-1/2013≠0 )`

`=>x=-2014`

a)

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\\ \Leftrightarrow2^x.1+2^x.2+2^x.2^2+2^x.2^3=120\\ \Leftrightarrow2^x\left(1+2+2^2+2^3\right)=120\\ \Leftrightarrow2^x=8=2^3\\ \Rightarrow x=3\)

b)

\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\\ \Leftrightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\\ \Leftrightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}=\dfrac{x+2015}{2013}+\dfrac{x+2015}{2014}\\ \Leftrightarrow\left(x+2015\right).\dfrac{1}{2011}+\left(x+2015\right).\dfrac{1}{2012}-\left(x+2015\right).\dfrac{1}{2013}-\left(x+2015\right).\dfrac{1}{2014}=0\\ \Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\\ \Rightarrow x+2015=0\Leftrightarrow x=-2015\)

đề câu b thiếu à bn

\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)

\(\Rightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\)

\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)

\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)

Mà \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\ne0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

Vậy x = -2015

\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)

\(\Rightarrow\dfrac{x+4}{2011}+\dfrac{x+3}{2012}-\dfrac{x+2}{2013}-\dfrac{x+1}{2014}=0\)

\(\Rightarrow\)\(\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)-\left(\dfrac{x+2}{2013}+1\right)-\left(\dfrac{x+1}{2014}+1\right)=0\)\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)

\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\) 

=> \(x+2014=0\) 

                  \(x=0-2014\) 

                  \(x=-2014\)

\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)

\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)

\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)

\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)

\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)