Nếu muốn trất thì: \(8\left(x-2015\right)^2;y^2\ge0\) và \(8\left(x-2015\right)^2⋮8\)

\(\Rightarrow\left\{{}\begin{matrix}8\left(x-2015\right)^2⋮8\\8\left(x-2015\right)^2\in N\le25\end{matrix}\right.\)

Y thì kệ nó tìm x rồi xử nó sau

x = 2015 , y = 5 nhé bn!

mik tính nhẩm thui!

Áp dụng tc dtsbn:

\(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}=\dfrac{x-z}{-2}=\dfrac{y-z}{-1}=\dfrac{x-y}{-1}\\ \Leftrightarrow\dfrac{x-z}{2}=\dfrac{y-z}{1}=\dfrac{x-y}{1}\\ \Leftrightarrow x-z=2\left(y-z\right)=2\left(x-y\right)\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

Ta có\(x\sqrt{\frac{\left(2015+y^2\right)\left(2015+z^2\right)}{2015+x^2}}=x\sqrt{\frac{\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}{xy+yz+zx+x^2}}\)

\(=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=xy+xz\)

Tương tự:\(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=yx+yz\)

               \(z\sqrt{\frac{\left(2015+x^2\right)\left(2015+y^2\right)}{2015+z^2}}=zx+zy\)

Ta có :\(P=xy+xz+yx+yz+zx+zy=2\left(xy+yz+zx\right)=4030\)

=>P không phải là số chính phương