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(2015x - 2014)³ = 8(x - 1)³ + (2013x - 2012)³

<=> 6(x - 1)(2013x - 2012)(2015x - 2014) = 0

Tới đây thì xong rồi

\(x^4-2015x^3+2015x^2-2015x+2015\)

\(=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)(vì x=2014 nên 2015=x+1)

\(=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(=1\)

nốt ý b:

\(\left(x-1\right)^3+1+3x\left(x-4\right)=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+1+3x^2-12x=0\)

\(\Leftrightarrow x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Vậy ..............

\(a,x\left(x-2012\right)-2013x+2012.2013=0\)

\(=x\left(x-2012\right)+2013\left(-x+2012\right)=0\)

\(\Rightarrow x\left(x-2012\right)-2013\left(x-2012\right)=0\)

\(\Rightarrow\left(x-2013\right)\left(x-2012\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2013=0\\x-2012=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2013\\x=2012\end{matrix}\right.\)

Vậy...

cho tui thì tui trả lời

Theo đề bài ta suy ra:

\(\left(x-2014\right)^3+\left(x+2012\right)^3=\left[2\left(x-1\right)\right]^3\Rightarrow\left(x-2014\right)^3+\left(x+2012\right)^3=\left(2x-2\right)^3\)(1)

Đặt \(\hept{\begin{cases}x-2014=a\\x+2012=b\end{cases}\Rightarrow}2x-2=a+b\)

Khi đó từ (1), ta có:

 \(a^3+b^3=\left(a+b\right)^3\Rightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\Rightarrow3ab\left(a+b\right)=0\)

\(\Rightarrow3\left(x-2014\right)\left(x+2012\right)\left(2x-2\right)=0\)

Từ đó tìm được \(x\in\left\{2014;-2012;1\right\}\)

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

-Bạn ỉn nè làm đúng r nhé !~

Học tốt 

nhé bạn ~

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

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

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

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

TH1:x -1 = 0

=>x=1

TH2 : x-2014=0

=> x=2014

\(x^3-4x=0\)

\(x\left(x^2-4\right)=0\)

\(x\left(x-4\right)\left(x+4\right)=0\)

TH1: x=0

TH2:x-4=0

=> x= 4

TH3: x+4=0

=> x=(-4)

Hok tốt

a=x⁴-2223x³+2223x²-2223x+2223

=x³(x-2223)+x(x-2223)+2222x²+2003(*)

thay x=2222,ta co:

(*)<=>-2222³-2222+2222³+2223=1

dung thi chon nha