\(\hept{\begin{cases}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{cases}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)

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

<=> x+y = 0 hoặc x+z=0 hoặc z+y=0

<=> x = -y hoặc x = -z hoặc z = -y

\(\Rightarrow P=\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)

\(\left\{{}\begin{matrix}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{matrix}\right.\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)

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

\(\Leftrightarrow x+y=0\) hoặc \(x+z=0\) hoặc \(z+y=0\)

\(\Leftrightarrow x=-y\) hoặc \(x=-z\) hoặc z=-y

\(\Rightarrow P\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)

Chúc bạn học tốt !!

\(x^2+y^2+z^2=xy+yz+zx\) và \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)

Ta có:

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

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

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

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

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Dấu " = " xảy ra :

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Thay \(x=y=z\) vào \(x^{2009}+y^{2009}+z^{2009}=3^{2009}\) ta được:

\(3x^{2009}=3x^{2010}\)

\(\Rightarrow x^{2009}=3^{2009}\)

\(\Rightarrow x=3\)

\(\Rightarrow y=z=x=3\)

Vậy \(\left(x;y;z\right)=\left(3;3;3\right)\)

Thiếu đề chăng.?

ta có: \(x^2+y^2\ge2xy\)

áp dụng tương tự cho với y,z và z,x

ta CM được: \(x^2+y^2+z^2\ge xy+yz+zx\)

Dấu = xaye ra <=> x=y=z

Thay vào pt 2 ta được: \(3x^{2009}=3^{2010}\Leftrightarrow x=3\)

vậy x=y=z=3

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

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)

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

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

\(\Leftrightarrow.....\)

dễ quá phát ơi

cho mình 9 k đúng đi phát

Lời giải:

Ta có:

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

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

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

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

Vì bản thân \((x-y)^2; (y-z)^2; (z-x)^2\geq 0, \forall x,y,z\in\mathbb{R}\) nên để tổng của chúng bằng $0$ thì \((x-y)^2=(y-z)^2=(z-x)^2=0\Rightarrow x=y=z\)

Khi đó:

\(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)

\(\Leftrightarrow 3x^{2009}=3y^{2009}=3z^{2009}=3^{2010}\Rightarrow x=y=z=3\)

Vậy........