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

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

a, Bạn cộng mỗi tỉ số với 1 rồi chuyển vế phải sang vế trái, ta được:

(x+2016)(1/2011 +1/2012 -1/2013 -1/2014) =0

Ta thấy thừa số thứ hai lớn hơn 0 nên x+2016=0

Vậy x=-2016

b, Bạn chuyển vế phải sang vế trái, ta có:

(5x-1,45)(1/6 +1/7 +1/8 -1/9 +1/10)=0

Thừa số thứ 2 lớn hơn 0 do đó:  5x -1,45 =0

                                                5x =1,45

                                                 x =0,29

Vậy x =0,29

Mong bạn hiểu cách giải của mình.

Chúc bạn học tốt.

Bạn xem lại đề nhé U Suck

Bài 3 : 

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

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

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

\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)

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

\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)

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

Nên \(x-2017=0\)

\(\Rightarrow\)\(x=2017\)

Vậy \(x=2017\)

Chúc bạn học tốt ~ 

Bài 1 : 

\(\left(8x-5\right)\left(x^2+2014\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)

Vậy \(x=\frac{5}{8}\)

Chúc bạn học tốt ~ 

Bài 2 : 

\(\frac{2-x}{2015}-1=\frac{1-x}{2016}-\frac{x}{2017}\)

\(\Leftrightarrow\)\(\frac{2-x}{2015}+1=\left(\frac{1-x}{2016}+1\right)-\left(\frac{x}{2017}-1\right)\) ( cộng 2 vế cho 2 ) 

\(\Leftrightarrow\)\(\frac{2-x+2015}{2015}=\frac{1-x+2016}{2016}-\frac{x-2017}{2017}\)

\(\Leftrightarrow\)\(\frac{2017-x}{2015}=\frac{2017-x}{2016}+\frac{2017-x}{2017}\)

\(\Leftrightarrow\)\(\frac{2017-x}{2015}-\frac{2017-x}{2016}-\frac{2017-x}{2017}=0\)

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

Vì \(\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)

Nên \(2017-x=0\)

\(\Rightarrow\)\(x=2017\)

Vậy \(x=2017\)

Chúc bạn học tốt ~ 

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

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

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

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

\(\Rightarrow x+2020=0\)

\(\Rightarrow x=-2020\)

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

a, \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

= \(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-1}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

( x + 5)(x - 3) = \(x^2-1\) - 8

x\(^2\) -3x + 5x -15 = \(x^2-9\)

= > \(x^2-x^2\) +2x = 15 - 9

=> 2x = 6

=> x = 3

bạn nói kết quả không thôi ư mình cần cả cách giải