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

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

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

=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0

=> x + 2017 = 0 

=> x = -2017

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

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

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

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

Dễ thấy cái ngoặc to < 0

=> x=-2017

(x-1)/2016 +(x-2)/2015 -(x-3)/2014 = (x-4)/2013.                 =>(x-1)/2016 +(x-2)/2015 = (x-3)/2014 + (x-4)/2013.      =>. (X-1)/2016 -1 + (x-2)/2015 -1 = (x -3)/2014 -1 + (x-4)/2013 -1 =>  (x -2017)/2016 + (x-2017)/2015 -(x-2017)/2014 -(x-2017)/2013 =0.   => (x-2017)(1/2016 +1/2015 -1/2014  -1/2013) = 0   => x-2017 =0 => x = 2017 

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

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

\(\Leftrightarrow\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)

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

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

Mà \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)

k cho mk

Đề bạn hình như hơi sai thì phải, nhưng nếu tìm x thì mình giải như sau

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

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

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

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

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

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

Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}< 0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

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

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

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

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

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

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

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

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

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

\(\Rightarrow x+2017=0\Rightarrow x=-2017\)

có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1

=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)

vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015

x=2015

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

\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)

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

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

\(x-2017=0\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)

\(x=2017\)

x-1/2016+x-2/2015-x-3/2014=x-4/2013

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

trừ mỗi phân số thêm 1 ta được

x-2017/2016+x-2017/2015-x-2017/2014-x-2017/2013=0

<=>(x-2017).(1/2016+1/2015+1/2014+1/2013)=0

(1/2016+1/2015+1/2014+1/2013) khác 0

=>x-2017=0

<=>x=2017

vậy x = 2017

X=2017

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

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

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

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

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

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

\(\Rightarrow x+2017=0\)

\(\Rightarrow x=-2017\)

Vậy \(x=-2017\)

bạn bấm vào "đúng 0" là sẽ có đáp án hiện ra

- 2017 nha bạn

x=-1

Ngắn gọn, k nhé