Đáp án:
e) Min=-1
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
P = \left[ {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{x\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x\left( {\sqrt x + 1} \right) + \left( {\sqrt x + 1} \right)}} + \dfrac{1}{{x + 1}}} \right]\\
= \left[ {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {x + 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{1}{{x + 1}}} \right]\\
= \left[ {\dfrac{{x + 1 - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x + 1}}{{x + 1}}} \right]\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)P < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - 2 - \sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{\sqrt x - 3}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \sqrt x - 3 < 0\\
\to x < 9\\
\to 0 \le x < 9;x \ne 1\\
c)P = \dfrac{1}{3}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{1}{3}\\
\to 3\sqrt x - 3 = \sqrt x + 1\\
\to 2\sqrt x = 4\\
\to x = 4\\
d)P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 2}}{{\sqrt x + 1}} = 1 - \dfrac{2}{{\sqrt x + 1}}\\
P \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 2\\
\sqrt x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = 0
\end{array} \right.\\
e)P = 1 - \dfrac{2}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 1 \ge 1\\
\to \dfrac{2}{{\sqrt x + 1}} \le 2\\
\to - \dfrac{2}{{\sqrt x + 1}} \ge - 2\\
\to 1 - \dfrac{2}{{\sqrt x + 1}} \ge - 1\\
\to Min = - 1\\
\Leftrightarrow x = 0
\end{array}\)
