\(\frac{x+4}{2013}\)+\(\frac{x+3}{2014}\)=\(\frac{x+2}{2015}\)+\(\frac{x+1}{2016}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HI
2
9 tháng 1 2020
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0
=> x + 2017 = 0
=> x = -2017
9 tháng 1 2020
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
\(\Leftrightarrow\frac{x+2}{2015}+1+\frac{x+1}{2016}+1=\frac{x+3}{2014}+1+\frac{x+4}{2013}+1\)
\(\Leftrightarrow\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Dễ thấy cái ngoặc to < 0
=> x=-2017
LE
1
LA
11 tháng 4 2019
\(\Leftrightarrow\left(\frac{x+4}{2013}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)
\(\Rightarrow x+2017=0\Rightarrow x=-2017\)
N
17 tháng 3 2019
Phải là \(\frac{x}{2012}\)
\(\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x=2012\)
Vậy ...
20 tháng 4 2018
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 ~
20 tháng 4 2018
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 ~
Y
1
24 tháng 1 2017
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 \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
27 tháng 9 2019
\(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+4}{2015}+\frac{x+5}{2014}+\frac{x+6}{2013}\)
\(\Leftrightarrow\) \(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{x+4}{2015}+1+\frac{x+5}{2014}+1+\frac{x+6}{2013}+1\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{x+2019}{2015}+\frac{x+2019}{2014}+\frac{x+2019}{2013}\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}-\frac{x+2019}{2015}-\frac{x+2019}{2014}-\frac{x+2019}{2013}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)\)\(=0\)
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
HP
13 tháng 12 2015
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
KN
11 tháng 2 2020
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
11 tháng 2 2020
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
16 tháng 2 2020
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
D
12 tháng 4 2016
\(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(x-2017=0\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)
\(x=2017\)
12 tháng 4 2016
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
\(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016} \)
\(\Leftrightarrow\frac{x+4}{2013}+1+\frac{x+3}{2014}+1-\frac{x+2}{2015}-1-\frac{x+1}{2016}-1=0\)
\(\Leftrightarrow\frac{x+2014+2013}{2013}+\frac{x+3+2014}{2014}-\frac{x+2+2015}{2015}-\frac{x+1+2016}{2016}=0\)
\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2017=0\) ( vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\)>0)
\(\Leftrightarrow x=2017\)
thank nha
>3