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![](https://rs.olm.vn/images/avt/0.png?1311)
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|x+1| + |x+2| + |x+3| + .......... + |x+2014| = 2015x
Ta có :
|x+1| \(\ge\)0
|x+2| \(\ge\)0
|x+3| \(\ge\)0
..........
|x+2014| \(\ge\)0
=> |x+1| + |x+2| + |x+3| +..........+ |x+2014| \(\ge\)0
=> 2015x \(\ge\)0
Mà 2015 \(\ge\)0
=> x \(\ge\)0
=> |x+1| + |x+2| + |x+3| +..........+ |x+2014|
= x + 1 + x + 2 + x + 3 +.................... + x + 2014 = 2015x
=> 2014x + (1 + 2 + 3 +............ + 2014) = 2015x
=> 1 + 2 + 3 + 4 + ........................ + 2014 = x
=> x = 2029105
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy :
\(\left|x+1\right|\ge0\)
\(\left|x+2\right|\ge0\)
............
|x + 2014| \(\ge0\)
Cộng vế với vế ta được :
\(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|\ge0\)
Mà \(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|=2015x\Rightarrow2015x\ge0\Rightarrow x\ge0\)\(\Rightarrow x+1+x+2+....+x+2014=2015x\)
\(\Rightarrow2014x+\frac{2014.2015}{2}=2015x\)
\(\Rightarrow2014x+2029105=2015x\)
\(\Rightarrow2015x-2014x=2029105\)
\(\Rightarrow x=2029105\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left|x+1\right|\ge0,\left|x+2\right|\ge0,...,\left|x+2014\right|\ge0\)
\(\Rightarrow\)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2014\right|\ge0\)
\(\Rightarrow2015x\ge0\)
\(\Rightarrow x\ge0\)
Khi đó :\(\left|x+1\right|=x+1,\left|x+2\right|=x+2,...,\left|x+2014\right|=x+2014\)\(\Rightarrow x+1+x+2+...+x+2014=2015x\)
\(\Rightarrow2014x+1+2+...+2014=2015x\)
\(\Rightarrow1+2+..+2014=x\)
\(\Rightarrow x=\dfrac{\left(1+2014\right)2014}{2}=2029105\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P(x) = x2016 - 2015x2015 - 2015x2014 - ... - 2015x2 - 2015x
<=> P(x) = x2016 - 2016x2015 + x2015 - 2016x2014 + x2014 - ... - 2016x2 + x2 - 2016x + x
<=> P(2016) = 20162016 - 2016.20162015 + 20162015 - 2016.20162014 + 20162014 -...- 2016.20162 + 20162 - 2016.2016 + 2016
<=> P(2016)=20162016 - 20162016 + 20162015 - 20162015 + 20162014 - ... - 20163 + 20162 - 20162 + 2016
<=> P(2016) = 2016
Vậy P(2016) = 2016
Ta có:
P(2016) = 20162016 - 2015 . 20162015 - 2015 . 20162014 -.....- 2015 . 20162 - 2015 . 2016 - 1
P(2016) = 20162016 - ( 2016 - 1 ) . 20162015 - ( 2016 -1 ) . 20162014 - ..... - ( 2016 - 1 ) . 20162 - ( 2016 - 1 ) . 2016 - 1
P(2016)= 20162016 - 20162016 + 20162015 - 20162015 + 20162014 - ..... - 20163 + 20162 - 20162 + 2016 - 1
P(2016) = 2016 - 1
P(2016) = 2015.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2014\right|\ge0\)
\(\Rightarrow2015x\ge0\Leftrightarrow x\ge0\)
Mà \(x\ge0\) nên ta có phương trình tương đương
\(x+1+x+2+x+3+...+x+2014=2015x\)
\(\Rightarrow\left(1+2+3+...+2014\right)+2014x=2015x\)
\(\Rightarrow x=1+2+3+...+2014=\frac{2014\left(2014+1\right)}{2}=2029105\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(|2015x-2014|=|2015x+2014|\)
\(\Leftrightarrow\orbr{\begin{cases}-2015x+2014=|2015x+2014|\left(l\right)\\2015x-2014=|2015x+2014|\left(n\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2015x+2014=-2015x+2014\\2015x+2014=2015x-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}4030x=0\\0x=-4028\left(l\right)\end{cases}\Leftrightarrow}4030x=0\Leftrightarrow x=0}\)
=> x là số nguyên dượng
=> x+1+x+2+x+3+x+4+...+x+2014=|x+1|+|x+2|+|x+3|+|x+4|+...+|x+2014|=2015
=> 2014x + (1+2+3+4+...+2014)=2015x
=> 1+2+3+4+...+2014=x