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:

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

Phân tích nhân tử là được

\(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=-z\\z=-x\end{cases}}\)

Với \(x=-y\) thì

\(\hept{\begin{cases}x^{2013}+y^{2013}+z^{2013}=z^{2013}\\\left(x+y+z\right)^{2013}=z^{2013}\end{cases}}\)

\(\Rightarrow x^{2013}+y^{2013}+z^{2013}=\left(x+y+z\right)^{2013}\)

Tương tự cho các trường hợp còn lại.

bt làm rồi hỏi vui thôi ^^

chỗ kia bạn ghi sai đề r:

mình sửa luôn

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

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

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

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

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

từ (1) và (2)

=> x=-y

nên

\(x^{2013}=-y^{2013}\) hay

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

\(2013.y+y.\frac{1}{2013}-2013=\frac{1}{2013}\)

\(\Rightarrow2013.y+y.\frac{1}{2013}=\frac{1}{2013}+2013\)

\(\Rightarrow y.\left(2013+\frac{1}{2013}\right)=2013+\frac{1}{2013}\)

\(\Rightarrow y=1\)

a: =>x^2+y^2+z^2-4x+2y-6z+14=0

=>x^2-4x+4+y^2+2y+1+z^2-6z+9=0

=>(x-2)^2+(y+1)^2+(z-3)^2=0

=>x=2; y=-1; z=3

b: \(\left(x+y+z\right)\cdot\left(xy+yz+xz\right)\)

\(=x^2y+xyz+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2\)

\(=x^2y+xy^2+y^2z+x^2z+yz^2+xz^2+3xyz\)

Theo đề, ta có:

\(x^2y+xy^2+y^2z+x^2z+yz^2+xz^2+2xyz=0\)

\(\Leftrightarrow x^2y+2xyz+yz^2+xy^2+2xzy+xz^2+zx^2-2xyz+zy^2=0\)

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

=>x=y=z=0

=>x^2013+y^2013+z^2013=(x+y+z)^2013

\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz+z+1}{xz+z+1}=1\)

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

Thay xyz=2013 vào ta có:

\(\frac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)

\(=\frac{xy\cdot xz}{xy\left(xz+z+1\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz+1+z}{xz+z+1}=1\) (Đpcm)