Bài làm

( x - 2014 )³ + ( x + 2012 )³ = 8( x - 1 )³ 

<=> ( x - 2014 )³ + ( x + 2012 )³ = 2³( x - 1 )³ 

<=> x- 2014 + x + 2012 = 2( x - 1 )

<=> 2x - 2 = 2x - 1

<=> 2x - 2x = 2 - 1

<=> 0x = 1

<=> x = 1 : 0 ( Vô lí )

Vậy pt trên vô nghiệm

<=> x = -4025 : 2

<=> x = 2012,5

Trl

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Học tốt 

nhé bạn ~

Hai dòng cuối bỏ nhé. Mik lm nhầm

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....

\(\Leftrightarrow\left(x-2014+x+2012\right)^3-3\cdot\left(x-2014\right)\left(x+2012\right)\left(x-2014+x+2012\right)=\left(2x-2\right)^3\)

=>3(x-2014)(x+2012)(2x-2)=0

=>\(x\in\left\{2014;-2012;1\right\}\)

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