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\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2^2-2.1=2\) (đpcm)
Do \(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)
=> \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=y+\dfrac{1}{z}\Leftrightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\Leftrightarrow x-y=\dfrac{y-z}{yz}\\y+\dfrac{1}{z}=z+\dfrac{1}{x}\Leftrightarrow y-z=\dfrac{1}{x}-\dfrac{1}{z}\Leftrightarrow y-z=\dfrac{z-x}{xz}\\z+\dfrac{1}{x}=x+\dfrac{1}{y}\Leftrightarrow z-x=\dfrac{1}{y}-\dfrac{1}{x}\Leftrightarrow z-x=\dfrac{x-y}{xy}\end{matrix}\right.\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\dfrac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)x^2y^2z^2=\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x^2y^2z^2-1\right)=0\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\) hoặc \(x^2y^2z^2-1=0\)
=> x=y=z hoặc xyz=1 hoặc xyz=-1
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)
\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+3xyz-xyz=0\)
\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz=0\)
\(\Leftrightarrow x^2y+xy^2+x^2z+xyz+y^2z+yz^2+xz^2+xyz=0\)
\(\Leftrightarrow x\left(xy+y^2+xz+yz\right)+z\left(y^2+yz+xz+xy\right)=0\)
\(\Leftrightarrow x\left[y\left(x+y\right)+z\left(x+y\right)\right]+z\left[y\left(y+z\right)+x\left(y+z\right)\right]=0\)
\(\Leftrightarrow x\left(x+y\right)\left(y+z\right)+z\left(y+z\right)\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
* x = -y
\(\dfrac{1}{x^{2007}}+\dfrac{1}{y^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{x^{2007}}-\dfrac{1}{x^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{z^{2007}}\)(*)
\(\dfrac{1}{x^{2007}+y^{2007}+z^{2007}}=\dfrac{1}{x^{2007}-x^{2007}+z^{2007}}=\dfrac{1}{z^{2007}}\)(*)
Từ (*) và (**) \(\Rightarrow\) đpcm
Tương tự xét y = -z và z = -x
Vậy nếu x, y, z khác 0 và x + y +z khác 0 thì \(\dfrac{1}{x^{2007}}+\dfrac{1}{y^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{x^{2007}+y^{2007}+z^{2007}}\).
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
