Đặt: \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+2bc+2ac+2c^2=0\)
\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\Leftrightarrow a=b=-c\Leftrightarrow x=y=-z\)
Giải tiếp nhé
\(\left(\frac{1}{x}+\frac{1}{y}\right)^2=\left(2-\frac{1}{z}\right)^2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}-\frac{1}{z^2}=4-\frac{4}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\Rightarrow\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=2\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}=-8\)
\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}-2=0\\\frac{1}{y}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\end{matrix}\right.\) \(\frac{1}{z}=2-\left(\frac{1}{x}+\frac{1}{y}\right)=-2\Rightarrow z=\frac{-1}{2}\)
Vậy \(P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2012}=1^{2012}=1\)
