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\(\Leftrightarrow\frac{x^{2014}}{a^2+b^2+c^2+d^2}+\frac{y^{2014}}{a^2+b^2+c^2+d^2}+\frac{z^{2014}}{a^2+b^2+c^2+d^2}+\frac{t^{2014}}{a^2+b^2+c^2+d^2}\)
\(-\frac{x^{2014}}{a^2}-\frac{y^{2014}}{b^2}-\frac{z^{2014}}{c^2}-\frac{t^{2014}}{d^2}=0\)
\(\Leftrightarrow\left(\frac{x^{2014}}{a^2+b^2+c^2+d^2}-\frac{x^{2014}}{a^2}\right)+\left(\frac{y^{2014}}{a^2+b^2+c^2+d^2}-\frac{y^{2014}}{b^2}\right)+\left(\frac{z^{2014}}{a^2+b^2+c^2+d^2}-\frac{z^{2014}}{c^2}\right)\)
\(+\left(\frac{t^{2014}}{a^2+b^2+c^2+d^2}-\frac{t^{2014}}{d^2}\right)=0\)
\(\Leftrightarrow x^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+y^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\right)+\)
\(z^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\right)+t^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\right)=0\)
vì a2,b2,c2,d2 lớn hơn hoặc bằng 0
=> \(\hept{\begin{cases}\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\ne0\end{cases}}và....\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\ne0\)
\(\Rightarrow\hept{\begin{cases}x^{2014}=0\\y^{2014}=0\\z^{2014}=0\end{cases}}và..t^{2014}=0\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}và...t=0\)
=> \(\hept{\begin{cases}x^{2015}=0\\y^{2015}=0\\z^{2015}=0\end{cases}}và..t^{2015}=0\Rightarrow x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)
vậy \(x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)
Ta có : \(\frac{x-2015}{2015}=\frac{y-2014}{2014}\)
=> \(\frac{x}{2015}-\frac{2015}{2015}=\frac{y}{2014}-\frac{2014}{2014}\)
=> \(\frac{x}{2015}-1=\frac{y}{2014}-1\)
=> \(\frac{x}{2015}=\frac{y}{2014}\)
=> \(\frac{x}{y}=\frac{2015}{2014}\)
Théo đề ta có (x-2015)/2015=(y-2014)/2014=x/2015-1=y/2014-1
--> x/2015=y/2014
-->x/y=2015/2014
tick nha
=>(x-2015).2014 = (y-2014).2015
=>2014x - 2014.2015 =2015y-2014.2015
=>2014x=2015y
=>\(\frac{x}{y}=\frac{2015}{2014}\)
=> (x-2015).2014 = (y-2014).2015
=> 2014x-2015.2014 = 2015y - 2014.2015
=> 2014x = 2015y
=> x/y=2015/2014
=> (x-2015).2014 = (y-2014).2015
=> 2014x-2015.2014 = 2015y - 2014.2015
=> 2014x = 2015y
=> x/y=2015/2014