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CM
0
CM
0
30 tháng 6 2020
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
16 tháng 7 2017
Ta có:
\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2014}-1\right)+\left(\dfrac{x+1}{2015}-1\right)+\left(\dfrac{x+2}{2016}-1\right)+\left(\dfrac{x+3}{2017}-1\right)+\left(\dfrac{x+4}{2018}-1\right)=0\)\(\Leftrightarrow\dfrac{x-2014}{2014}+\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}+\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}=0\)\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)=0\) (1)
Mà \(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}>0\) (2)
Từ (1) và (2) => \(x-2014=0\) \(\Leftrightarrow x=2014\)
4 tháng 11 2016
Điều kiện x \(\ge0\)từ đó ta có
x + 2015 + x + 2016 + x + 2017 = 6x
<=> x = 2016
4 tháng 11 2016
Ta có \(\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|\ge0\Rightarrow6x\ge0\Rightarrow x\ge0\)
=> \(\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|=3x+6048=6x\Rightarrow3x=6048\Rightarrow x=2016\)
Vậy x=2016
4 tháng 11 2016
Ta có:
\(\left|x+2015\right|\ge0\)
\(\left|x+2016\right|\ge0\)
\(\left|x+2017\right|\ge0\)
\(\Rightarrow\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x\right|=x\)
\(\Rightarrow\left|x+2015\right|+\left|x+2016\right|+\left|x+2017\right|=\left(x+2015\right)+\left(x+2016\right)+\left(x+2017\right)=6x\)
\(\Rightarrow3x+6048=6x\)
\(\Rightarrow3x=6048\)
\(\Rightarrow x=2016\)
Vậy \(x=2016\)
DV
2
21 tháng 3 2020
\(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}\)
\(\Leftrightarrow\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}-\frac{x+2015}{5}-\frac{x+2016}{4}-\frac{x+2017}{3}=0\)
\(\Leftrightarrow\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)-\left(\frac{x+2015}{5}+1\right)-\left(\frac{x+2016}{4}+1\right)\)
\(-\left(\frac{x+2017}{3}+1\right)=0\)
\(\Leftrightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{5}-\frac{x+2020}{4}-\frac{x+2020}{3}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x+2020=0\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
<=> x=-2020
Vậy x=-2020
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
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
vì x-2017>=0
=>x-2014+x-2015+x-2016=x-2017
<=>2x=4028
<=>x=2014