Đáp án:
\(\begin{array}{l}
6)\dfrac{1}{A} < 2\\
7)Max = - 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 1\\
A = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + {{\left( {\sqrt x - 1} \right)}^2} + 3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 + x - 2\sqrt x + 1 + 3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2x + 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {2\sqrt x + 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x - 1}}\\
6)\dfrac{1}{A} - 2 = \dfrac{{\sqrt x - 1}}{{2\sqrt x + 1}} - 2\\
= \dfrac{{\sqrt x - 1 - 4\sqrt x - 2}}{{2\sqrt x + 1}}\\
= \dfrac{{ - 3 - 3\sqrt x }}{{2\sqrt x + 1}}\\
= - \dfrac{{3\sqrt x + 3}}{{2\sqrt x + 1}}\\
Do:x \ge 0 \to \left\{ \begin{array}{l}
3\sqrt x + 3 > 0\\
2\sqrt x + 1 > 0
\end{array} \right.\\
\to \dfrac{{3\sqrt x + 3}}{{2\sqrt x + 1}} > 0\\
\to - \dfrac{{3\sqrt x + 3}}{{2\sqrt x + 1}} < 0\\
\to \dfrac{1}{A} - 2 < 0\\
\to \dfrac{1}{A} < 2\\
7)A.\left( {1 - x} \right) = \dfrac{{2\sqrt x + 1}}{{\sqrt x - 1}}.\left( {1 - \sqrt x } \right)\left( {\sqrt x + 1} \right)\\
= - \left( {2\sqrt x + 1} \right)\left( {\sqrt x + 1} \right)\\
= - \left( {2x + 3\sqrt x + 1} \right)\\
= - \left( {2x + 2.\sqrt {2x} .\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8} - \dfrac{1}{8}} \right)\\
= - {\left( {\sqrt {2x} + \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{1}{8}\\
Do:\sqrt {2x} \ge 0\forall x \ge 0\\
\to \sqrt {2x} + \dfrac{3}{{2\sqrt 2 }} \ge \dfrac{3}{{2\sqrt 2 }}\\
\to {\left( {\sqrt {2x} + \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge \dfrac{9}{8}\\
\to - {\left( {\sqrt {2x} + \dfrac{3}{{2\sqrt 2 }}} \right)^2} \le - \dfrac{9}{8}\\
\to - {\left( {\sqrt {2x} + \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{1}{8} \le - 1\\
\to Max = - 1\\
\Leftrightarrow x = 0
\end{array}\)
