ai giups tui di

Nhân cả 2 vế của đẳng thức đã cho với \(\left(x-\sqrt{x^2+\sqrt{2015}}\right)\)ta được:

\(-\sqrt{2015}\left(y+\sqrt{y^2+\sqrt{2015}}\right)=\sqrt{2015}\left(x-\sqrt{x^2+\sqrt{2015}}\right)\)(1)

Nhân cả 2 vế của đẳng thức đã cho với \(\left(y-\sqrt{y^2+\sqrt{2015}}\right)\)ta được:

\(-\sqrt{2015}\left(x+\sqrt{x^2+\sqrt{2015}}\right)=\sqrt{2015}\left(y-\sqrt{y^2+\sqrt{2015}}\right)\)(2)

Lấy (1) + (2), ta được:

\(-\sqrt{2015}\left(y+\sqrt{y^2+\sqrt{2015}}+x+\sqrt{x^2+\sqrt{2015}}\right)\)

\(=\sqrt{2015}\left(x+y-\sqrt{x^2+\sqrt{2015}}-\sqrt{y^2+\sqrt{2015}}\right)\)

\(\Leftrightarrow x+y+\sqrt{x^2+\sqrt{2015}}+\sqrt{y^2+\sqrt{2015}}\)

\(=-x-y+\sqrt{x^2+\sqrt{2015}}+\sqrt{y^2+\sqrt{2015}}\)

\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)

Vậy x + y = 0

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

\(\left(x+\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\)

\(\Leftrightarrow\frac{2015}{\sqrt{x^2+2015}-x}\left(y+\sqrt{y^2+2015}\right)=2015\)

\(\Leftrightarrow\sqrt{x^2+2015}-x=y+\sqrt{y^2+2015}\left(1\right)\)

Tương tự : \(x+\sqrt{x^2+2015}=\sqrt{y^2+2015}-y\left(2\right)\)

(1)+(2):

\(x+\sqrt{x^2+2015}+y+\sqrt{y^2+2015}=\sqrt{x^2+2015}+\sqrt{y^2+2015}-x-y\)

\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)

Ta có:\(\left(x+\sqrt{x^2+2015}\right)\left(y-\sqrt{y^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\left(y-\sqrt{y^2+2015}\right)\)

\(\Leftrightarrow-2015\left(x+\sqrt{x^2+2015}\right)=2015\left(y-\sqrt{y^2+2015}\right)\)

\(\Leftrightarrow x+\sqrt{x^2+2015}=\sqrt{y^2+2015}-y\)                                                 (1)

 Lại có:\(\left(x+\sqrt{x^2+2015}\right)\left(x-\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\left(x-\sqrt{x^2+2015}\right)\)

\(\Leftrightarrow-2015\left(y+\sqrt{y^2+2015}\right)=2015\left(x-\sqrt{x^2+2015}\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2015}=\sqrt{x^2+2015}-x\)                                               (2)

Cộng theo vế \(\left(1\right)\) và \(\left(2\right)\) ta có:\(x+\sqrt{x^2+2015}+y+\sqrt{y^2+2015}=\sqrt{y^2+2015}+\sqrt{x^2+2015}-x-y\)

\(\Leftrightarrow2x+2y=0\Leftrightarrow x+y=0\)

Từ \(\left(x+\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)=2015\)

\(\Leftrightarrow2015\left(x+\sqrt{x^2+2015}\right)=2015\left(\sqrt{y^2+2015}-y\right)\)

\(\Leftrightarrow x+\sqrt{x^2+2015}=\sqrt{y^2+2015}-y\left(1\right)\)

Tương tự cũng có: \(y+\sqrt{y^2+2015}=\sqrt{x^2+2015}-x\left(2\right)\)

\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)

Từ giả thuyết ta đc x+y=0 thì =>x^2015+y^2015=(x+y)(...)=0

cái đoạn x+y=0 bạn xem mấy bài đăng khác ấy!:>>

Nhân liên hợp zô :v

\(\Rightarrow\left(x+\sqrt{x^2+2015}\right)\left(y+\sqrt{y^2+2015}\right)\left(x-\sqrt{x^2+2015}\right)=2015\left(x-\sqrt{x^2+2015}\right)\)

\(\Leftrightarrow-2015\left(y+\sqrt{y^2+2015}\right)=2015\left(x-\sqrt{x^2+2015}\right)\)

\(\Leftrightarrow-\left(y+\sqrt{y^2+2015}\right)=x-\sqrt{x^2+2015}\)

Tương tự, ta cũng suy ra:

\(-\left(x+\sqrt{x^2+2015}\right)=y-\sqrt{y^2+2015}\)

Cộng vế theo vế ta được x+y=0

Ahihi :v

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