Đáp án:
1) \(P = \dfrac{{3 + \sqrt x }}{{\sqrt x - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0;x \ne 1\\
A = \dfrac{6}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{\sqrt x - 1}} = \dfrac{{6 + \sqrt x }}{{\sqrt x - 1}}\\
B = \dfrac{3}{{\sqrt x - 1}}\\
P = A - B = \dfrac{{6 + \sqrt x }}{{\sqrt x - 1}} - \dfrac{3}{{\sqrt x - 1}}\\
= \dfrac{{3 + \sqrt x }}{{\sqrt x - 1}}\\
2)\dfrac{1}{P} = \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 - 4}}{{\sqrt x + 3}} = 1 - \dfrac{4}{{\sqrt x + 3}}\\
Do:\sqrt x \ge 0 \to \sqrt x + 3 \ge 3\\
\to \dfrac{4}{{\sqrt x + 3}} \le \dfrac{4}{3}\\
\to - \dfrac{4}{{\sqrt x + 3}} \ge - \dfrac{4}{3}\\
\to 1 - \dfrac{4}{{\sqrt x + 3}} \ge 1 - \dfrac{4}{3}\\
\to 1 - \dfrac{4}{{\sqrt x + 3}} \ge - \dfrac{1}{3}\\
\to Min = - \dfrac{1}{3}\\
\Leftrightarrow x = 0
\end{array}\)
