Ta có: 25 - y² = 8(x-2014)²

y²=25-8.(x-2014)²

Vì y² là số dương và x,y thuộc N

=>(x-2014)²<=3

Xét (x-2014)²=0

=>x=2014

=>y=5

Xét (x-2014)²=1

=> x=2015 hoặc x=2013

Không tìm được y thỏa mãn

Vậy x=2014;y=5

Mk làm tắt đấy.Sai thì thôi nhé

\(x^2+y^2+z^2=xy+yz+zx\)

\(2.\left(x^2+y^2+z^2\right)=2.\left(xy+yz+zx\right)\)

\(\Rightarrow2.\left(x^2+y^2+z^2\right)-2xy-2yz-2zx=0\)

\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Ta có: \(VT\ge0\forall x;y;z\)( tự c/m. nếu b ko c/m được thì bảo mình )

Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow}}\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)

Có \(x^{2014}+y^{2014}+z^{2014}=3\)

\(\Rightarrow3.x^{2014}=3\)

\(\Rightarrow x^{2014}=1\)

\(\Rightarrow x=1\)

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

Có: \(P=x^{25}+y^4+z^{2015}\)

\(\Rightarrow P=1^{25}+1^4+1^{2015}\)

\(P=1+1+1\)

\(P=3\)

Vậy \(P=3\)

Tham khảo nhé~

Ta có: x²+y²+z²=xy+yz+zx

<=>2x²+2y²+2z²=2xy+2yz+2zx

<=>2x²+2y²+2z²-2xy-2yz-2zx=0

<=>(x²-2xy+y²)+(y²-2yz+z²)+(z²-2zx+x²)=0

<=>(x-y)²+(y-z)²+(z-x)²=0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)

=>\(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow x=y=z}\)

=>x²⁰¹⁴=y²⁰¹⁴=z²⁰¹⁴

Lại có: x²⁰¹⁴+y²⁰¹⁴+z²⁰¹⁴ = 3

=>3x²⁰¹⁴ = 3 => x²⁰¹⁴ = 1 => \(x=\pm1\)

=>\(x=y=z=\pm1\)

Thay x,y,z vào P rồi tính

Nhầm.

Tui thiếu trường hợp x=-1

b tham khảo bài của  ST nhé

Vì \(8\left(x-2009\right)^2\) chẵn nên \(25-y^2\) chẵn

Mà \(25\) lẻ nên \(y^2\) lẻ

Và \(25-y^2=8\left(x-2009\right)^2\ge0\Leftrightarrow y^2\le25\)

\(\Leftrightarrow y^2\in\left\{1;9;25\right\}\Leftrightarrow y\in\left\{1;3;5\right\}\left(y\in N\right)\)

\(\forall y=1\Leftrightarrow8\left(x-2009\right)^2=24\Leftrightarrow\left(x-2009\right)^2=3\left(loại\right)\\ \forall y=3\Leftrightarrow8\left(x-2009\right)^2=16\Leftrightarrow\left(x-2009\right)^2=2\left(loại\right)\\ \forall y=5\Leftrightarrow8\left(x-2009\right)^2=0\Leftrightarrow\left(x-2009\right)^2=0\Leftrightarrow x=2009\left(nhận\right)\)

Vậy \(\left(x;y\right)=\left(2009;5\right)\)

Ta co : 8(x-2014)² = 25-y²

=> 8(x-2014)² + y² = 25 ^(*)

Voi moi \(y\in N\) ta co y² \(\ge0\)

\(\Rightarrow8\left(x-2014\right)^2\le25\)

\(\Rightarrow\left(x-2014\right)^2\le\dfrac{25}{3}\)

Vi x\(\in N\)

\(\Rightarrow\left(x-2014\right)^2=0hoac\left(x-2014\right)^2=1\)

Neu\(\left(x-2014\right)^2=1\) thay vao(*) ta duoc;

8 . 1+ y² =25

\(\Rightarrow25-8=y^2\)

17 = y² (loai) (vi y \(\in N\))

Neu \(\left(x-2014\right)^2=0\) thay vao (*) ta duoc:

8 . 0 + y² = 25

=> y² = 25

=> y = 5 (vi y\(\in N\))

Khi do \(\left(x-2014\right)^2=0\)

=> x- 2014 = 0 => x = 2014

Vay x = 2014, y = 5

\(\left(x-2014\right)^2=\frac{25-y^2}{8}\)

\(x=\sqrt{\frac{25-y^2}{8}}+2014\)

Để x nguyên \(\Leftrightarrow\sqrt{\frac{25-y^2}{8}}\in N\forall y\in[-5;5]\)

\(\Rightarrow\left\{{}\begin{matrix}x=2014\\y=5\end{matrix}\right.\)