<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)

\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)

=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)

=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)

vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)

=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0

=> \(x+4025=0\)

\(x=-4025\)

khì +1 vào mỗi phân số      

pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0

<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0

<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0

<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )

<=> x=2012

Vậy x=2012

Tk mk nha

Ta có : 

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

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

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

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

Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)

\(\Rightarrow\)\(x-2012=0\)

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

Vậy \(x=2012\)

Chúc bạn học tốt ~

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

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

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

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

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

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

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

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

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

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

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

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

1.

<=> B=\(3^{24}-\left\lbrack\left(3^3\right)^4+1\right\rbrack\left\lbrack\left(3^2\right)^6-1\right\rbrack\)

\(B=3^{24}-\left(3^{12}+1\right)\left(3^{12}-1\right)\)

\(B=3^{24}-3^{24}+1\)

\(B=1\)

2.

xét vế đầu tiên

\(2011\cdot2013+2012\cdot2014\)

\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)

\(=2012^2-1+2013^2-1\)

\(=2012^2+2013^2-2\)

=> \(2011\cdot2013+2012\cdot2014=2012^2+2013^2-2\)

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

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

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

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

\(\Leftrightarrow x=2010\)