cộng 1 vào mỗi tỉ số ta được:

\(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1+\frac{x+3}{2014}+1=\frac{x+4}{2013}+1+\frac{x+5}{2012}+\frac{x+6}{2011}\)

=>\(\frac{x+1}{2016}+\frac{2016}{2016}+\frac{x+2}{2015}+\frac{2015}{2015}+\frac{x+3}{2014}+\frac{2014}{2014}=\frac{x+4}{2013}+\frac{2013}{2013}+\frac{x+5}{2012}+\frac{2012}{2012}+\frac{x+6}{2011}+\frac{2011}{2011}\)

=>

\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}=\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\)

=>

\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\left(\frac{x+2017}{2013}+\frac{x+2017}{2012}+\frac{x+2017}{2011}\right)=0\)

=>

\(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)

=>(x+2017).(1/1016+1/2015+1/2014-1/2013-1/2012-1/2011)=0

dễ thấy 1/2016<1/2015<1/2014<1/2013<1/2012<1/2011

=>1/2016+...-1/2011 khác 0

=>x+2017=0

=>x=-2017

nhớ tick

Ta có : 

\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+8}{2010}+\frac{x+7}{2011}\)

\(\Leftrightarrow\)\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+4+2014}{2014}+\frac{x+3+2015}{2015}=\frac{x+8+2010}{2010}+\frac{x+7+2011}{2011}\)

\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)

\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\ne0\)

Nên \(x-2018=0\)

\(\Leftrightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~ 

Ta có: \(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+7}{2011}+\frac{x+8}{2010}\)

\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+7}{2011}+1\right)+\left(\frac{x+8}{2010}+1\right)\)

\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}=\frac{x+2018}{2011}+\frac{x+2018}{2010}\)

\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}-\frac{x+2018}{2011}-\frac{x+2018}{2010}=0\)

\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2018=0\Rightarrow x=-2018\)

Chúc bn hc tốt! ^_^

\(\Leftrightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1=\frac{x+8}{2010}+1+\frac{x+7}{2011}+1\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Do \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\ne0\Rightarrow x+2018=0\Leftrightarrow x=-2018\)
 

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 \(x-2011=0\)

\(x=2011\)

Bài 1 : 

Ta có  : 

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

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

\(+\left(\frac{x+2013}{2011}+1\right)\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)

\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)

\(\Rightarrow x+4024=0\)

\(\Rightarrow x=-4024\)

Bài 2 : 

Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)

=> Phương trình trở thành 

\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)

\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)

\(\Rightarrow5a^2+3a-8=0\)

\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)

Vì \(a\ge0\Rightarrow a=1\)

\(\Rightarrow x^2+2x+1=1\)

\(x^2+2x=0\)

\(\Rightarrow x\left(x+2\right)=0\)

\(\Rightarrow x\in\left\{-2,0\right\}\)

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

Vậy x = -10

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

\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)

\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

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

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

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

Vậy x = 2012

a, (x+1)/9 +1 + (x+2)/8  =  (x+3)/7 + 1 + (x+4)/6 + 1

<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6

<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0

vì 1/9 +1/8 -1/7 - 1/6 khác 0

=> x+10=0

=> x=-10

\(\frac{x+5}{2015}+\frac{x+6}{2014}+\frac{x+7}{2013}+\frac{x+8}{2012}+\frac{x+9}{2011}+5=0\)

\(\Rightarrow1+\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1+\frac{x+8}{2012}+1+\frac{x+9}{2011}=0\)

\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}+\frac{x+2020}{2012}+\frac{x+2020}{2011}=0\)

\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)

\(\Rightarrow x+2020=0\)

\(\Rightarrow x=-2020\)

Study well 

chuyên toán thcs

Thiếu giải thích

Trả lời 

không cần giải thích nha 

\(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}=0\right)\)

\(\Rightarrow\left(x+2020\right)=0:\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)

\(\Rightarrow\left(x+2020\right)=0\)

Tôi ngại chuyển vế nên làm tắt có cần giả thích cái j đâu