=> 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

Chính là 0

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

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

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

|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

[ ] là giá trị tuyệt đối à

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....

\(x^2-2015x+2014=0\)

\(x^2-2014x-x+2014=0\)

\(x\left(x-2014\right)-\left(x-2014\right)=0\)

\(\left(x-2014\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2014=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2014\\x=1\end{cases}}}\)

\(x^2-2015x+2014\)\(=0\)

\(\Rightarrow x^2-x-2014x+2014\)\(=0\)

\(\Rightarrow x\left(x-1\right)-2014\left(x-1\right)\)\(=0\)

\(\Rightarrow\left(x-1\right)\left(x-2014\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2014=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2014\end{cases}}\)

thanks