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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....
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em gửi bài qua fb thầy chữa cho, tìm fb của thầy bằng sđt nhé: 0975705122
(2015x - 2014)3 = 8(x - 1)3 + (2013x - 2012)3
<=> 6(x - 1)(2013x - 2012)(2015x - 2014) = 0
Tới đây thì xong rồi
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\(x^4-2015x^3+2015x^2-2015x+2015\)
\(=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)(vì x=2014 nên 2015=x+1)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(=1\)
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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...
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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 \(x\in\left\{2014;-2012;1\right\}\)
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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
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\(x^2-2015x+2014=0\)
\(x^2-x-2014x+2014=0\)
\(x\left(x-1\right)-2014\left(x-1\right)=0\)
\(\left(x-1\right)\left(x-2014\right)=0\)
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
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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