Ta có: \(\sqrt{x^2 + 2013}>\vert{}x\vert{}\ge x\) ∀x∈R

=>\(\sqrt{x^2 + 2013}-x>0\)

\(\left(\sqrt{x^2 + 2013} - x\right)\left(\sqrt{x^2 + 2013} + x\right) = (x^2 + 2013) - x^2 = 2013\)

Ta có: \((x + \sqrt{x^2 + 2013})(y + \sqrt{y^2 + 2013}) = 2013\)

=>\(\left(\sqrt{x^2 + 2013} - x\right)\left(\sqrt{x^2 + 2013} + x\right)\left(\sqrt{y^2 + 2013} + y\right) = 2013 \left(\sqrt{x^2 + 2013} - x\right)\)

=>\(2013\left(\sqrt{y^2 + 2013}+y\right)=2013\left(\sqrt{x^2 + 2013}-x\right)\)

=>\(\sqrt{y^2+2013}+y=\sqrt{x^2 + 2013}-x\quad(1)\)

Ta có: \((x + \sqrt{x^2 + 2013})(y + \sqrt{y^2 + 2013}) = 2013\)

=>\(\left(\sqrt{x^2 + 2013} + x\right)\left(\sqrt{y^2 + 2013} - y\right)\left(\sqrt{y^2 + 2013} + y\right) = 2013 \left(\sqrt{y^2 + 2013} - y\right)\)

=>\(2013\left(\sqrt{x^2 + 2013}+x\right)=2013\left(\sqrt{y^2 + 2013}-y\right)\)

=>\(\sqrt{x^2 + 2013} + x = \sqrt{y^2 + 2013} - y \quad (2)\)

Từ (1) và (2) suy ra \(\left(\sqrt{y^2 + 2013} + y\right) + \left(\sqrt{x^2 + 2013} + x\right) = \left(\sqrt{x^2 + 2013} - x\right) + \left(\sqrt{y^2 + 2013} - y\right)\)

=>x+y=-x-y

=>x+y=0

=>x=-y

\(x^{2013}+y^{2013}=\left(-y\right)^{2013}+y^{2013}=0\)

Ta có:

\(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\\ \Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\\ \Leftrightarrow y+\sqrt{y^2+2013}=\sqrt{x^2+2013}-x\left(1\right)\)

Tương tự: \(x+\sqrt{x^2+2013}=\sqrt{y^2+2013}-y\left(2\right)\)

Do đó: 2x=-2y

Suy ra: x=-y

Do đó:

\(x^{2013}+y^{2013}=\left(-y\right)^{2013}+y^{2013}=0\left(ĐPCM\right)\)

Bạn xem lại đề nhé.

a) \(A=x^2+5y^2+2xy-4x-8y+2015\)

\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2-y\right)^2+4y^2+2011\)

Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)

\(\Rightarrow A_{min}=2011\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

Bạn quy đồng rồi phân tích tử thành nhân tử rồi ra à.

Xét x=2012 hoặc x=2013 thì thỏa mãn pt 

Xét x>2013 hoặc x<2012 thì pt vô nghiệm 

Xét 2013>x>2012 thì \(\left|x-2012\right|^{2013}< \left|x-2012\right|=x-2012\)

                                    \(\left|x-2013\right|^{2012}< \left|x-2013\right|=2013-x\)

Cộng vào => VT<VP => vô lí 

Vậy ...

^^

X=2013

Ta có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)mà xy+yz+zx=0

\(\Rightarrow x^2+y^2+z^2=0\left(1\right)\)

Lại có: \(x^2,y^2,z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)Kết hợp (1)

\(\Leftrightarrow x^2=y^2=z^2=0\Leftrightarrow x=y=z=0\)

Vậy \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=-1+0+1=0\)

Ta có : \(x+y+z=0\)

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

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

\(\Rightarrow x^2+y^2+z^2=0\) ( Do \(xy+yz+zx=0\) )

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

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

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

\(\Rightarrow x=y=z\)

Khi đó : \(x+y+z=3x=0\)

\(\Rightarrow x=0\Rightarrow x=y=z=0\)

Nên \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=0\)

Vậy : \(T=0\).