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1/x + 1/y + 1/z = 1/x+y+z
<=> xy+yz+zx/xyz = 1/x+y+z
<=> (xy+yz+xz).(x+y+z)=xyz
<=> x^2y+xy^2+y^2z+z^2y+z^2x+x^2z+3xyz=xyz
<=> x^2y+y^2x+y^2z+z^2y+z^2x+x^2z+2xyz = 0
<=> (x+y).(y+z).(z+x) = 0
<=> x+y=0 hoặc y+z=0 hoặc x+z=0
<=> x=-y hoặc y=-z hoặc z=-x
Nếu x=-y => x^25 = -y^25 => P = 0
Nếu y=-z => y^3 = -z^3 => P = 0
Nếu z=-x => z^2006 = x^2006 => P = 0
Vậy P = 0
Tk mk nha
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\frac{xy+z\left(x+y+z\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
Vậy x+y=0, y+z=0 hoặc z+x=0
TH1: Nếu x+y=0 => \(x=-y\Rightarrow x^{25}+y^{25}=0\Rightarrow P=0\)
TH2: Nếu y+z=0 => \(y=-z\Rightarrow y^3+z^3=0\Rightarrow P=0\)
TH3: Nếu z+x=0 => \(z=-z\Leftrightarrow z^{2006}-x^{2006}=0\Rightarrow P=0\)
Vậy P=0
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)=1\)
\(\Leftrightarrow3xyz+yz\left(y+z\right)+xz\left(x+z\right)+xy\left(x+y\right)=xyz\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) hay B = 0
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
\(\Rightarrow A=0\)
Ta có :\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=x^2+y^2+z^2=0\) (do xy + yz + xz = 0)
Ta lại thấy \(x^2;y^2;z^2\ge0\forall x;y;z\) nên \(x^2+y^2+z^2\ge0\forall x;y;z\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=0\) thay vào S ta được :
\(S=\left(-1\right)^{2005}+\left(-1\right)^{2006}+1^{2007}=1\)