Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow\left(\frac{x+4}{2013}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
\(\Rightarrow x+2017=0\Rightarrow x=-2017\)
\(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(x-2017=0\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)
\(x=2017\)
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
\(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}\)
\(\Rightarrow\frac{x+4}{2013}+1+\frac{x+3}{2014}+1=\frac{x+2}{2015}+1+\frac{x+1}{2016}+1\)
\(\Rightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}=\frac{x+2017}{2015}+\frac{x+2017}{2016}\)
\(\Rightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
\(Do\)\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
\(\Rightarrow x+2017=0\)
\(\Rightarrow x=-2017\)
Vậy \(x=-2017\)
bạn bấm vào "đúng 0" là sẽ có đáp án hiện ra
(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 \(x-2017=0\Leftrightarrow x=2017\)
\(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)
\(\Leftrightarrow\)\(\frac{x+2}{2013}+1+\frac{x+1}{2014}+1=\frac{x}{2015}+1+\frac{x-1}{2016}+1\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2014}=\frac{x+2015}{2015}+\frac{x+2015}{2016}\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Do\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}>0\)
=>x+2015=0
<=>x=-2015
=> \(\frac{x+2015-2013}{2013}+\frac{x+2015-2014}{2014}=\frac{x+2015-2015}{2015}+\frac{x+2015-2016}{2016}\)
<=> \(\frac{x+2015}{2013}-1+\frac{x+2015}{2014}-1=\frac{x+2015}{2015}-1+\frac{x+2015}{2016}-1\)
<=> \(\frac{x+2015}{2013}+\frac{x+2015}{2014}-\frac{x+2015}{2015}-\frac{x+2015}{2016}=0\)
<=> \(\left(x+2015\right).\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
<=> x + 2015 = 0 Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
<=> x = -2015
\(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+4}{2015}+\frac{x+5}{2014}+\frac{x+6}{2013}\)
\(\Leftrightarrow\) \(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{x+4}{2015}+1+\frac{x+5}{2014}+1+\frac{x+6}{2013}+1\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{x+2019}{2015}+\frac{x+2019}{2014}+\frac{x+2019}{2013}\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}-\frac{x+2019}{2015}-\frac{x+2019}{2014}-\frac{x+2019}{2013}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)\)\(=0\)
Lại có: \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\) \(\ne\) \(0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=0-2019=-2019\)
Vậy x= -2019
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 \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
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
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\left(\frac{x-1}{2016}+1\right)+\left(\frac{x-2}{2015}+1\right)=\left(\frac{x-3}{2014}+1\right)+\left(\frac{x-4}{2013}+1\right)\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2014}\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x+2017}{2014}-\frac{x+2017}{2013}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
=> (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