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:

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

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

Theo đề bài  và (1) => dpcm

b) theo a có \(y+\sqrt{y^2+\sqrt{2013}}=-x+\sqrt{x^2+\sqrt{2013}}\)(2)

tương tự ta có \(x+\sqrt{x^2+\sqrt{2013}}=-y+\sqrt{y^2+\sqrt{2013}}\)(3)

Cộng 2 vế (2)  với (3) => x+y = -x -y

hay 2(x+y) =0 =>S= x+y =0

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

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

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

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

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

Chứng minh tương tự: \(x+\sqrt{x^2+2013}=-y+\sqrt{y^2+2013}\)

cộng vế theo vế ta được: \(x+y=-x-y\)

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

\(\Leftrightarrow x^{2013}+y^{2013}=0\)

a,Ta có x =...

x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)}{\left(\sqrt{\sqrt{3}+1}\right)\left(\sqrt{\sqrt{3}-1}\right)}\)

x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}\)

x = \(\frac{\sqrt{3}.2}{\sqrt{3}}\)

x = 2

sau đó thay x=2 vào A nhé.

A=2014 !!!

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

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

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

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

Lấy (1) - (2) ta được -2x = 2y

                          <=> 2x + 2y = 0

                          <=> P = x + y = 0

Dễ dàng nhận ra \(x-\sqrt{x^2+2013}\ne0\), nhân 2 vế với nó:

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

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

Tương tự ta có \(x+\sqrt{x^2+2013}=\sqrt{y^2+2013}-y\)

Cộng vế với vế:

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

\(\Rightarrow2\left(x+y\right)=0\Rightarrow P=0\)

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

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

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

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

=> x=-y

=> x+y = 0

Tk mk nha

          \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2103\)

Ta có:   \(\hept{\begin{cases}\left(x+\sqrt{x^2+2013}\right)\left(\sqrt{x^2+2013}-x\right)=2013\\\left(y+\sqrt{y^2+2013}\right)\left(\sqrt{y^2+2013}-y\right)=2013\end{cases}}\)

Kết hợp với giả thiết ta có:

\(\hept{\begin{cases}\sqrt{x^2+2013}-x=y+\sqrt{y^2+2013}\\\sqrt{y^2+2013}-y=x+\sqrt{x^2+2013}\end{cases}}\)

Cộng  theo vế ta có:    \(-x-y=x+y\)

\(\Rightarrow\)\(A=x+y=0\)