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\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
Có : \(\left|x-2014\right|+\left|x-2015\right|+\left|x-2016\right|=\left|x-2014\right|+\left|2016-x\right|+\left|x-2015\right|\)
\(\ge\left|x-2014+2016-x\right|+\left|x-2015\right|\)
\(=2+\left|x-2015\right|\ge2\)
Dấu "=" khi x = 2015
Vậy x = 2015
\(\left|x-2014\right|+\left|x-2015\right|+\left|x-2016\right|=2\)
\(\Leftrightarrow\left|x-2014\right|+\left|2016-x\right|+\left|x-2015\right|=2\)
\(\Leftrightarrow\left|x-2014+2016-x\right|+\left|x-2015\right|\ge2\)
\(\Leftrightarrow2+\left|x-2015\right|\ge2\)
Dấu "=" xảy ra:
\(\Leftrightarrow2+\left|x-2015\right|=2\)
\(\Leftrightarrow x-2015=0\)
\(\Leftrightarrow x=2015\)
Vậy biểu thức trên = 2 khi x = 2015
Ta có:\(\left|x-2013\right|+\left|x-2014\right|+\left|y-2015\right|+\left|x-2016\right|\)
\(=\left|x-2013\right|+\left|2016-x\right|+\left|x-2014\right|+\left|y-2015\right|\)
\(\ge\left|x-2013+2016-x\right|+\left|x-2014\right|+\left|y-2015\right|\)
\(=3+\left|x-2014\right|+\left|y-2015\right|\)
\(\ge3+0+0=3\)
Mà \(\left|x-2013\right|+\left|x-2014\right|+\left|y-2015\right|+\left|x-2016\right|=3\)
\(\Rightarrow\) Dấu "=" xảy ra khi và chỉ khi:
\(\hept{\begin{cases}\left(x-2013\right)\left(2016-x\right)\ge0\\\left|x-2014\right|=0\\\left|y-2015\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2013\le x\le2016\left(1\right)\\x=2014\left(2\right)\\y=2015\end{cases}}\)
Dễ thấy \(\left(2\right)\) thỏa mãn \(\left(1\right)\) nên \(x=2014;y=2015\)
\(\frac{x+2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)
\(\Leftrightarrow\frac{x+2015}{2016}+1+\frac{x+2016}{2015}+1+\frac{x+2017}{2014}+1=0\)
\(\Leftrightarrow\frac{x+4031}{2016}+\frac{x+4031}{2015}+\frac{x+4031}{2014}=0\)
\(\Leftrightarrow\left(x+4031\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
\(\Rightarrow x+4031=0\)
\(\Rightarrow x=-4031\)
Lập bảng xét dấu
Ta được 4 trường hợp sau:
-Nếu \(x< 2014\) thì \(\left|x-2014\right|+\left|x-2015\right|+\left|x-2016\right|=3\) (1)
\(\Leftrightarrow2014-x+2015-x+2016-x=3\)
\(\Leftrightarrow6045-3x=3\)
\(\Leftrightarrow3x=6042\)
\(\Leftrightarrow x=2014\) (loại)
-Nếu \(2014\le x< 2015\) thì (1) tương đương:
\(x-2014+2015-x+2016-x=3\)
\(\Leftrightarrow2017-x=3\)
\(\Leftrightarrow x=2014\) (nhận)
-Nếu \(2015\le x< 2016\) thì (1) tương đương:
\(x-2014+x-2015+2016-x=3\)
\(\Leftrightarrow-2013+x=3\)
\(\Leftrightarrow x=2016\) (loại)
-Nếu \(x\ge2016\) thì (1) tương đương:
\(x-2014+x-2015+x-2016=3\)
\(\Leftrightarrow-6045+3x=3\)
\(\Leftrightarrow3x=6048\)
\(\Leftrightarrow x=2016\) (nhận)
Vậy x = 2014 hoặc x = 2016