=> 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

|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

toán bồi dưởng à

Ko bài kiểm tra

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 :\(\left|x+1\right|=x+1,\left|x+2\right|=x+2,...,\left|x+2014\right|=x+2014\)\(\Rightarrow x+1+x+2+...+x+2014=2015x\)

\(\Rightarrow2014x+1+2+...+2014=2015x\)

\(\Rightarrow1+2+..+2014=x\)

\(\Rightarrow x=\dfrac{\left(1+2014\right)2014}{2}=2029105\)

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.

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

\(x+1+x+2+x+3+...+x+2014=2015x\)

\(\Rightarrow\left(1+2+3+...+2014\right)+2014x=2015x\)

\(\Rightarrow x=1+2+3+...+2014=\frac{2014\left(2014+1\right)}{2}=2029105\)

Chính là 0

\(|2015x-2014|=|2015x+2014|\)

\(\Leftrightarrow\orbr{\begin{cases}-2015x+2014=|2015x+2014|\left(l\right)\\2015x-2014=|2015x+2014|\left(n\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2015x+2014=-2015x+2014\\2015x+2014=2015x-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}4030x=0\\0x=-4028\left(l\right)\end{cases}\Leftrightarrow}4030x=0\Leftrightarrow x=0}\)