Đáp án đúng:
Giải chi tiết:\(\begin{array}{l}\,\,x + y = 3 \Rightarrow y = 3 - x\\ \Rightarrow xy = x\left( {3 - x} \right) = - {x^2} + 3x = - {x^2} + 3x - 2 + 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {x - 1} \right)\left( {2 - x} \right) + 2 \le 2\,\,\,\forall \,\,0 < x \le 1.\\P = \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{3}{{xy}} \ge \frac{3}{2}.\end{array}\)
Dấu bằng xảy ra:
\( \Leftrightarrow \left[ \begin{array}{l}x - 1 = 0\\2 - x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\,\,\,\left( {tm} \right)\\x = 2\,\,\,\left( {ktm} \right)\end{array} \right. \Rightarrow y = 3 - x = 3 - 1 = 2\,\,\,\left( {tm} \right).\)
Vậy \(Min\,\,\,P=\frac{3}{2}\,\,\,khi\,\,\,\left\{ \begin{align} & x=1 \\ & y=2 \\ \end{align} \right..\)
