x=1006

z=1007

y= -1005

Ta có: \(\frac{x+y}{2014}\)=\(\frac{x-y}{2016}\)

=>\(2016x+2016y=2014x-2014y\)

=> \(2x=-4030y\)

=>\(x=-2015y\)

\(Thay\)\(x=-2015\)vào \(\frac{x+y}{2014}=\frac{xy}{2015}\)ta được

\(\frac{-2015+y}{2014}=\frac{-2015y}{2015}\)

\(\frac{-2014y}{2014}=\frac{-2015y^2}{2015}\)

\(-y=-y^2\)

=>\(y-y^2=0\)

   \(y\).(\(1-y\))\(=0\)

\(=>\orbr{\begin{cases}y=0\\1-y=0\end{cases}}=>\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

TH1 :\(y=0=>x.y=-2015.0=0\)

TH2 :\(y=1=>x.y=-2015.1=-2015\)

x=0 y=0

x=-2015 y=1

Ta có: \(\frac{x+y}{2014}\ne\frac{x-y}{2016}\)

\(\Leftrightarrow2016x+2016y=2014x-2014y\)

\(\Leftrightarrow2x=-4030y\)

\(\Leftrightarrow x=-2015y\)

Thay \(x=-2015y\)vào \(\frac{x+y}{2014}=\frac{xy}{2015}\)ta được:

\(\Leftrightarrow\frac{-2015+y}{2014}=\frac{-2015y}{2015}\)

\(\Leftrightarrow\frac{-2014y}{2014}=\frac{-2015y^2}{2015}\)

\(\Leftrightarrow-y=-y^2\)

\(\Leftrightarrow y-y^2=0\)

\(\Leftrightarrow y\left(1-y\right)=0\)

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

Trường hợp \(y=0\):

\(y=0\Rightarrow x.y=-2015.0=0\)

Trường hợp \(y=1\):

\(y=1\Rightarrow x.y=-2015.1=-2015\)

hình như phải là dấu = bn ak

x=y=0

mà z đâu ra thế

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x+y}{2012}=\dfrac{xy}{2013}=\dfrac{x-y}{2014}=\dfrac{x+y+x-y}{2012+2014}=\dfrac{2x}{4026}=\dfrac{x}{2013}\)

\(hay:\dfrac{xy}{2013}=\dfrac{x}{2013}\Rightarrow xy=x\Rightarrow y=1\)

Ta có:

\(\dfrac{x+y}{2012}=\dfrac{x}{2013}\) (c/m trên)

\(\Rightarrow\left(x+y\right)2013=2012x\\ hay:\left(x+1\right)2013=2012x\\ \Rightarrow2013x+2013=2012x\\ \Rightarrow2013x-2012x=-2013\\ \Rightarrow x=-2013\)

Vậy: x=-2013

=>(x-1)/2015 - 1 + (x-2(/2014 -1 = (x-3)/2013 -1 + (x-4)/2012 -1

=>(x-2016)*(1/2015+1/2014-1/2013-1/2012)=0

=>x=2016

Trừ 1 ở mỗi p/s,ta có:

\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)=\left(\frac{x-3}{2013}-1\right)+\left(\frac{x-4}{2012}-1\right)\)

\(\Leftrightarrow\left(\frac{x-2016}{2015}\right)+\left(\frac{x-2016}{2014}\right)=\left(\frac{x-2016}{2013}\right)+\left(\frac{x-2016}{2012}\right)\)

\(\Leftrightarrow\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

Vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)

=>x-2016=0

=>x=2016

Vậy..................

mình hỏi bài này :tìm số tự nhiên x biết :(x-2)^2014=(x-2)^2016

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

x-y/2014=x+y/2012=x-y+x+y/2014+2012=2x/2026=x/1013 (theo tc dãy tỉ số bằng nhau)

ta lại có: x+y/2012=x.y/2013=x/2013 (chứng minh trên) => y=1

x-y/2014=x.y/2013=x/2013 => x-1/2014=x/2013

thì (x-1).2013=2014x

    2013x-2013=2014x

   -1x=2013 thì x=-2013

\(\Rightarrow x^{2014}+y^{2014}-2\left(x^{2013}+y^{2013}\right)+x^{2012}+y^{2012}=0\)

\(\Leftrightarrow x^{2012}.\left(x-1\right)^2+y^{2012}.\left(y-1\right)^2=0\)

\(\Rightarrow x=1;y=1\)

\(\Rightarrow P=2\)

cai gi

hgjkgj