=> 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

toán bồi dưởng à

Ko bài kiểm tra

[ ] là giá trị tuyệt đối à

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 \(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|=2015x\Rightarrow2015x\ge0\Rightarrow x\ge0\)\(\Rightarrow x+1+x+2+....+x+2014=2015x\)

\(\Rightarrow2014x+\frac{2014.2015}{2}=2015x\)

\(\Rightarrow2014x+2029105=2015x\)

\(\Rightarrow2015x-2014x=2029105\)

\(\Rightarrow x=2029105\)

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

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

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

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

PT <=> (2015x - 2014)³ = (2x - 2)³ + (2013x - 2012)³

<=> (2015x - 2014)³ = (2x - 2 + 2013x - 2012). [(2x-2)² - (2x - 2).(2013x - 2012) + (2013x - 2012)²]

<=>    (2015x - 2014)³ = (2015x - 2014). [(2x-2)² - (2x - 2).(2013x - 2012) + (2013x - 2012)²]

<=> (2015x - 2014).[ (2015x - 2014)² -  [(2x-2)² - (2x - 2).(2013x - 2012) + (2013x - 2012)²]] = 0 

<=> 2015.x - 2014 = 0 hoặc (2015x - 2014)² -  [(2x-2)² - (2x - 2).(2013x - 2012) + (2013x - 2012)²] = 0 

+) 2015x - 2014 = 0 => x = 2014/2015

+) (2015x - 2014)² -  [(2x-2)² - (2x - 2).(2013x - 2012) + (2013x - 2012)²] = 0

<=> [(2x - 2) + (2013x - 2012)]² - (2x - 2)² + (2x - 2).(2013x - 2012) - (2013x - 2012)² = 0 

<=> 3. (2x - 2).(2013x - 2012) = 0 

<=> 2x - 2 = 0 hoặc 2013x - 2012 = 0 

<=> x = 1 hoặc x = 2012/2013

Vậy....