Đáp án:
\(\min P = 7 + 2\sqrt6 \Leftrightarrow x = \dfrac{2}{6+\sqrt6}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad P = \dfrac{1}{1-3x} + \dfrac{2}{x}\qquad \qquad\left(0 < x < \dfrac12\right)\\
\to P = \dfrac{1}{1-3x} + \dfrac{6}{3x}\\
\to P \geqslant \dfrac{(1+\sqrt6)^2}{1 - 3x + 3x}\quad (BDT\ Cauchy-Schwarz\ dạng\ Engel)\\
\to P \geqslant (1 + \sqrt6)^2\\
\to P \geqslant 7 + 2\sqrt6\\
\text{Dấu = xảy ra}\ \Leftrightarrow \dfrac{1 - 3x}{1} = \dfrac{3x}{\sqrt6} \Leftrightarrow x = \dfrac{2}{6 + \sqrt6}\\
\text{Vậy}\ \min P = 7 + 2\sqrt6 \Leftrightarrow x = \dfrac{2}{6+\sqrt6}
\end{array}\)
