Đáp án:
$min_P=2 \Leftrightarrow x=4.$
Giải thích các bước giải:
$\sqrt{P}=\sqrt{\dfrac{x}{\sqrt{x}-1}}\\ \text{ĐKXĐ:} \left\{\begin{array}{l} x \ge 0 \\ \dfrac{x}{\sqrt{x}-1} \ge 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}-1> 0(\text{Do }x \ge 0)\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}> 1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x> 1\end{array} \right.\\ \Leftrightarrow x> 1\\ \sqrt{P}=\sqrt{\dfrac{x}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{x-2\sqrt{x}+1+2\sqrt{x}-2+1}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{(\sqrt{x}-1)^2+2(\sqrt{x}-1)+1}{\sqrt{x}-1}}\\ =\sqrt{(\sqrt{x}-1)+\dfrac{1}{\sqrt{x}-1}+2}\\ \ge \sqrt{2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}+2}(Cauchy)\\ =2$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}-1=\dfrac{1}{\sqrt{x}-1}$
$\Leftrightarrow (\sqrt{x}-1)^2=1\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-1=1 \\ \sqrt{x}-1=-1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}=2 \\ \sqrt{x}=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=4 \\ x=0(L)\end{array} \right.$
Vậy $min_P=2 \Leftrightarrow x=4.$
