Bạn tham khảo nhé : 

Ta có : 

\(\frac{x-3}{2012}+\frac{x-2}{2013}=\frac{x-2013}{2}+\frac{x-2012}{3}\)

\(\Leftrightarrow\)\(\left(\frac{x-3}{2012}-1\right)+\left(\frac{x-2}{2013}-1\right)=\left(\frac{x-2013}{2}-1\right)+\left(\frac{x-2012}{3}-1\right)\)

\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}=\frac{x-2015}{2}+\frac{x-2015}{3}\)

\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}-\frac{x-2015}{2}-\frac{x-2015}{3}=0\)

\(\Leftrightarrow\)\(\left(x-2015\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\right)=0\)

Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\ne0\)

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

\(\Rightarrow\)\(x=2015\)

Vậy \(x=2015\)

Chsuc bạn học tốt ~

chuyển vế t cs 

x+x-x-x=....

0x=.....

Suy ra vô nghiệm

(=) [(x-3/2012)-1]+[(x-2/2013)-1]= [(x-2013/2)-1]+[(x-2012/3)-1]

(=) x-2015/2012 .   +.   x-2015/2013. = x-2015/2.     +.     x-2015/3

(=)( x-2015 ) × (1/2012    + . 1/2013.    -. 1/2 .   -   1/3 .) =0

Mà 1/2012+1/2013-1/2-1/3≠0

=) x-2015 =0

(=) x=2015

ok

1)

\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)

\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)

\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)

\(\Leftrightarrow x=2015\)

Vậy \(S=\left\{2015\right\}\)

x = 2013

Câu 1 bị sai đề bài.

Câu 2:

\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)

\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}+\frac{1}{2011}\)

Vì:

\(\frac{1}{2011}>\frac{1}{2012};\frac{1}{2011}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}>0\)

\(\Rightarrow\)\(\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)

\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)

Tính 2 mũ76 - 2 mũ 74/ 2 mũ 78- 2 mũ 76

bài này khó lắm đó

x-y=2011

y-z=-2012

cộng 2 pt này ra đc : x-z=-1mà x+z=2013 .Công 2 pt này ta đc 2x=2012 hay x=1006

nên z=1007

Thay x=1006 vào x-y=2011 thì y=-1005

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 ~

2/Tham khảo: Câu hỏi của kudo shinichi - Toán lớp 8

 \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)

Đặt \(\hept{\begin{cases}2x^2+x-2013=m\\x^2-5x-2012=n\end{cases}}\)nên ta có phương trình:

\(m^2+4n^2=4nm\)

\(\Leftrightarrow m^2-2.m.2n+\left(2n\right)^2=0\)

\(\Leftrightarrow\left(m-2n\right)^2=0\)

Tự làm nốt...

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