Đáp án đúng: C
Giải chi tiết:Ta có: \({\left\{ \begin{array}{l}xy + yz + zx = 3z\\x + 2y = 3z\end{array} \right.^2} \Leftrightarrow {\left\{ \begin{array}{l}\frac{{xy + yz + zx}}{{{z^2}}} = 3\\\frac{x}{z} + 2\frac{y}{z} = 3\end{array} \right.^{}} \Leftrightarrow \left\{ \begin{array}{l}\frac{x}{z}.\frac{y}{z} + \frac{y}{z} + \frac{x}{z} = 3\,\,\,\,\,\left( {chia\,\,ca\,\,2\,\,ve\,\,cho\,\,{z^2}} \right)\\\frac{x}{z} + 2\frac{y}{z} = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {chia\,\,ca\,\,2\,\,ve\,\,cho\,\,z} \right)\end{array} \right.\)
Đặt \(\begin{array}{l}\left\{ \begin{array}{l}\frac{x}{z} = a > 0\\\frac{y}{z} = b > 0\end{array} \right.\\\end{array}\) ta có:
\(\begin{array}{l}\left\{ \begin{array}{l}\frac{x}{z}.\frac{y}{z} + \frac{y}{z} + \frac{x}{z} = 3\\\frac{x}{z} + 2\frac{y}{z} = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a + b + ab = 3\\a + 2b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3 - 2b + b + \left( { - 2b + 3} \right)b = 3\\a = - 2b + 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 2{b^2} + 2b = 0\\a = - 2b + 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}b = 0\,\,(loai)\\b = 1\end{array} \right.\\a = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\frac{x}{z} = 1\\\frac{y}{z} = 1\end{array} \right. \Leftrightarrow x = y = z\\\end{array}\)
Thay vào P ta có \(P = \frac{{x + 2z}}{y} = \frac{{3z}}{z} = 3\).
Vậy \(P = 3\).
