PT
2
K
Khách
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FZ
0
FZ
1
15 tháng 10 2015
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....
PT
1
18 tháng 11 2022
\(\Leftrightarrow\left(x-2014+x+2012\right)^3-3\cdot\left(x-2014\right)\left(x+2012\right)\left(x-2014+x+2012\right)=\left(2x-2\right)^3\)
=>3(x-2014)(x+2012)(2x-2)=0
=>\(x\in\left\{2014;-2012;1\right\}\)
PT
2
KT
25 tháng 4 2018
\(A=B\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}=-2^2\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x+3}{2013}+1+\frac{x+4}{2012}+1=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}+\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\)\(\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}\right)=0\)
\(\Leftrightarrow\)\(x+2016=0\) (do 1/2015 + 1/2014 + 1/2013 + 1/2012 # 0)
\(\Leftrightarrow\)\(x=-2016\)
Vậy...
20 tháng 4 2018
Bài 3 :
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-4}{2013}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1-2016}{2016}+\frac{x-2-2015}{2015}=\frac{x-3-2014}{2014}+\frac{x-4-2013}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
Nên \(x-2017=0\)
\(\Rightarrow\)\(x=2017\)
Vậy \(x=2017\)
Chúc bạn học tốt ~
20 tháng 4 2018
Bài 1 :
\(\left(8x-5\right)\left(x^2+2014\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{5}{8}\)
Chúc bạn học tốt ~
13 tháng 8 2016
1/ Ta có : \(P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\)
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
2/ Ta có : \(x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1\)
3/ \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-\frac{1}{2}\) \(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)(vì a+b+c=0)
Ta có : \(a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}\)
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 \(x\in\left\{2014;-2012;1\right\}\)