NL
2
K
Khách
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15 tháng 3 2016
(x+2/2014)+1 + (x+1/2015)+1 = (x+2016)+1 + (x-1/2017)+1
(x+2016/2014) + (x+2016/2015) - (x+2016/2016) - (x-2016/2017)=0
=>(x+2016)(1/2014+1/2015-1/2016-1/2017)
vì 1/2014+1/2015-1/2016-1/2017 luôn khác 0 => x+2016=0
=> x=-2016
HT
0
30 tháng 6 2020
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
D
2
27 tháng 2 2018
\(\frac{x-3}{2015}+\frac{x-2}{2016}=\frac{x-2016}{2}+\frac{x-2015}{3}\)
\(\Leftrightarrow\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-2}{2016}-1\right)=\left(\frac{x-2016}{2}-1\right)+\left(\frac{x-2015}{3}-1\right)\)
\(\frac{x-2018}{2015}+\frac{x-2018}{2016}-\frac{x-2018}{2}-\frac{x-2018}{3}=0\)
\(\Leftrightarrow\left(x-2018\right).\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}< 0\)
nên x - 2018 = 0
,<=> x = 2018
Vậy phương có 1 nghiệm là x = 2018
27 tháng 2 2018
pt <=> (x-3/2015 - 1) + (x-2/2016 - 1) = (x-2016/2 - 1) + (x-2015/3 - 1)
<=> x-2018/2015 + x-2018/2016 = x-2018/2 + x-2018/3
<=> x-2018/2 + x-2018/3 - x-2018/2015 - x-2018/2016 = 0
<=> (x-2018).(1/2+1/3-1/2015-1/2016) = 0
<=> x-2018 = 0 ( vì 1/2+1/3-1/2015-1/2016 > 0 )
<=> x=2018
Tk mk nha
16 tháng 11 2016
\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=2016-2016\)
\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)
\(\Rightarrow\left(x-2017\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\right)=0\)
Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)
=> x = 2017
NL
2
22 tháng 1 2019
\(\frac{x-1}{2015}+\frac{x-2}{2014}+...+\frac{x-2014}{2}+x=4030\)
\(\Leftrightarrow\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+...+\left(\frac{x-2014}{2}-1\right)+x-2016=0\)
\(\Leftrightarrow\frac{x-2016}{2015}+\frac{x-2016}{2014}+...+\frac{x-2016}{2}+x-2016=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}+1\right)=0\)
\(\Leftrightarrow x=2016\)
PD
1
6 tháng 3 2018
\(PT\Leftrightarrow\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=2017\)
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 ...
MA
23 tháng 9 2016
\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
HC
23 tháng 9 2016
<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)
<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)
<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)
<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)
<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)
<=> \(x=2017\)
Vậy x = 2017
đúng thì
\(\Leftrightarrow\frac{2-x}{2014}-1=-\frac{x+2012}{2014}\)
\(\Rightarrow\frac{1-x}{2015}-\frac{x}{2016}=-\frac{4031x-2016}{4062240}\)
\(\Rightarrow-\frac{x+2012}{2014}=-\frac{4031x-2016}{4062240}\)
\(\Rightarrow-\frac{x}{2014}-\frac{1006}{1007}=\frac{1}{2015}-\frac{4031x}{4062240}\)
\(\Rightarrow\frac{2028097x}{4090675680}-\frac{2028097}{2029105}=0\)
\(\Rightarrow\frac{2028097\left(x-2016\right)}{4090675680}=0\)
=>x=2016
\(\frac{2-x}{2014}-1=\frac{1-x}{2015}-\frac{x}{2016}\) \(\left(\text{*}\right)\)
Cộng hai vế của phương trình trên với \(2\) , khi đó, phương trình \(\left(\text{*}\right)\) trở thành:
\(\frac{2-x}{2014}+1=\left(\frac{1-x}{2015}+1\right)+\left(1-\frac{x}{2016}\right)\)
\(\Leftrightarrow\) \(\frac{2016-x}{2014}=\frac{2016-x}{2015}+\frac{2016-x}{2016}\)
\(\Leftrightarrow\) \(\left(2016-x\right)\left(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\) nên \(2016-x=0\) \(\Leftrightarrow\) \(x=2016\)
Vậy, tập nghiệm của pt \(\left(\text{*}\right)\) là \(S=\left\{2016\right\}\)