https://dethi.violet.vn/present/showprint/entry_id/11072330

bạn vào link trên sẽ có full đề và đáp án 

p/s: nhớ k cho mình nha <3

\(\frac{x-2}{4}=-\frac{16}{2-x}\)

\(\Leftrightarrow\frac{x-2}{4}=\frac{16}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=4.16=64\)

\(\Leftrightarrow\left(x-2\right)^2=8^2\)

\(\Leftrightarrow\left(x-2-8\right)\left(x-2+8\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+6\right)=0\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)

ta thấy x=0;y=0 là 1 nghiệm

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 \(y=1\):

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

a) 

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 \(y=1\):

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

Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành: 

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)

\(\Rightarrow x=2018;y=2019;z=2020\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)

\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)

\(x=2018,y=2019,z=2020\)

ĐK : \(\hept{\begin{cases}x>2014\\y>2015\\z>2016\end{cases}}\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2014}-1}{x-2014}+\frac{1}{4}-\frac{\sqrt{y-2015}-1}{y-2015}+\frac{1}{4}-\frac{\sqrt{z-2016}-1}{z-2016}=0\)

\(\Leftrightarrow\frac{x-2010-4\sqrt{x-2014}}{4\left(x-2014\right)}+\frac{y-2011-4\sqrt{y-2015}}{4\left(y-2015\right)}+\frac{z-2012-4\sqrt{z-2016}}{4\left(x-2014\right)}=0\)

\(\Leftrightarrow\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}+\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}+\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}=0\)( 1 )

Mà \(\hept{\begin{cases}\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}\ge0\forall x>2014\\\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}\ge0\forall y>2015\\\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}\ge0\forall z>2016\end{cases}}\)( 2 )

Từ ( 1 ) và ( 2 ) => \(\hept{\begin{cases}\left(2-\sqrt{x-2014}\right)^2=0\\\left(2-\sqrt{y-2015}\right)^2=0\\\left(2-\sqrt{z-2016}\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}\sqrt{x-2014}=2\\\sqrt{y-2015}=2\\\sqrt{z-2016}=2\end{cases}}\)<=>\(\hept{\begin{cases}x=2018\\y=2019\\z=2020\end{cases}}\)( tmđk )

Vậy ( x ; y ; z ) = ( 2018 ; 2019 ; 2020 )

Hoặc là sai đề hoặc là x,y,z đều bằng 0.

Đề bạn hình như hơi sai thì phải, nhưng nếu tìm x thì mình giải như sau

Ta có: \(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)

\(\Rightarrow\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-4}{2013}+\frac{x-3}{2014}\)

\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1=\frac{x-4}{2013}-1+\frac{x-3}{2014}-1\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2013}+\frac{x-2017}{2014}\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)

\(\Rightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)

Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}< 0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\left(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}\right)+\left(\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}\right)+\left(\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}\right)=0\)

\(\Leftrightarrow\left(x^2.\frac{b^2+c^2}{a^2+b^2+c^2}\right)+\left(y^2.\frac{a^2+c^2}{a^2+b^2+c^2}\right)+\left(z^2.\frac{a^2+b^2}{a^2+b^2+c^2}\right)=0\)

Vì a,b,c khác 

=>Dấu bằng xảy ra khi x=y=z=0

\(\Rightarrow x^{2014}+y^{2015}+z^{2016}=0^{2014}+0^{2015}+0^{2016}=0\)