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

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:

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

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

c/ ĐKXĐ: \(x\ge3\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x-3}-\sqrt{x-2}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x-1\right)\left(x+3\right)}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2< 3\left(ktm\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

a/ ĐKXĐ: \(\left\{{}\begin{matrix}x>2013\\y>2014\\z>2015\end{matrix}\right.\)

\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2013}-1}{x-2013}+\frac{1}{4}-\frac{\sqrt{y-2014}-1}{y-2014}+\frac{1}{4}-\frac{\sqrt{z-2015}-1}{z-2015}=0\)

\(\Leftrightarrow\frac{x-2013-4\sqrt{x-2013}+4}{4\left(x-2013\right)}+\frac{y-2014-4\sqrt{y-2014}+4}{4\left(y-2014\right)}+\frac{z-2015-4\sqrt{z-2015}+4}{4\left(z-2015\right)}=0\)

\(\Leftrightarrow\left(\frac{\sqrt{x-2013}-2}{2\sqrt{x-2013}}\right)^2+\left(\frac{\sqrt{y-2014}-2}{2\sqrt{y-2014}}\right)^2+\left(\frac{\sqrt{z-2015}-2}{2\sqrt{z-2015}}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2013}-2=0\\\sqrt{y-2014}-2=0\\\sqrt{z-2015}-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)

b/ Trừ vế cho vế 2 pt ta được:

\(x^3-y^3=2\left(y-x\right)\)

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

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

\(\Leftrightarrow\left(x-y\right)\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}+2\right]=0\)

\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)

Thay vào pt đầu:

\(x^3+1=2x\Leftrightarrow x^3-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow...\)

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

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 !!!

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:

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

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