Đề đúng phải là:

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)

Cộng mỗi phân thức thêm 1, quy đồng rồi chuyển sang 1 vế ta được:

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+2015}{2003}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Mà BT tích sau luôn nhỏ hơn 0

=> x+2015=0 => x = -2015

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)( như này đúng không ? :)) )

<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+10}{2005}+1\right)+\left(\frac{x+11}{2004}+1\right)+\left(\frac{x+12}{2003}+1\right)\)

<=> \(\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}=\frac{x+10+2005}{2005}+\frac{x+11+2004}{2004}+\frac{x+12+2003}{2003}\)

<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}=\frac{x+2015}{2005}+\frac{x+2015}{2004}+\frac{x+12}{2003}\)

<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+12}{2003}=0\)

<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)

=> x + 2015 = 0

=> x = -2015

a) (x-5)^x+2015 - (x-5)^x+2014  =0

  (x-5)^x+2014(x-5 -1) =0

+ x -5 =0 => x =5

+ x -6 =0 => x =6

Vậy x = 5 hoặc x =6

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

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

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

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

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

Vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\Rightarrow x-2016=0\)

\(\Rightarrow x=2016\)

b) \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)

\(\Rightarrow\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)

\(\Rightarrow\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)

\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)

\(\Rightarrow\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

vì \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\Rightarrow x-2005=0\)

\(\Rightarrow x=2005\)

c) \(|5x-3|\ge7\)

\(\Rightarrow5x-3\ge7\) hoặc - (5x-3) \(\ge7\)

\(\Rightarrow5x-3\ge7\) hoặc \(-5x+3\ge7\)

\(\Rightarrow5x\ge10\) hoặc \(-5x\ge4\)

\(\Rightarrow x\ge2\) hoặc \(x\le\frac{4}{-5}\)

k nhé!!! Kp luôn nha!

Ta có : \(\dfrac{x-2012}{8}+\dfrac{x-2008}{6}+\dfrac{x-2005}{5}=10-\dfrac{x-2004}{4}\)

\(\Leftrightarrow\left(\dfrac{x-2012}{8}-1\right)+\left(\dfrac{x-2008}{6}-2\right)+\left(\dfrac{x-2005}{5}-3\right)+\left(\dfrac{x-2004}{4}-4\right)=0\)\(\Leftrightarrow\dfrac{x-2020}{8}+\dfrac{x-2020}{6}+\dfrac{x-2020}{5}+\dfrac{x-2020}{4}=0\)

\(\Leftrightarrow\left(x-2020\right).\left(\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)

Vậy x = 2020

2)

đặt a= 1+2-3-4+5+6-........+2002-2003-2004+2005+2006

Biểu thức a có (2006-1)/1+1=2006(số hạng)

Nhóm 4 số hạng vào một nhóm ta có 2006 / 4= 501 dư 2 số hạng để ra một số đầu và một số cuối

a= 1+(2-3-4+5)+(6-7-8+9)-.........+(2002-2003-2004+2005) + 2006

a=1+0+0+......+0+2006

a=1+2006

a=2007 

vậy a = 2007