Đáp án:
b) \(\begin{array}{l}
TH1:B > 0\\
\Leftrightarrow x > 4\\
TH2:B < 0\\
\to 0 < x < 4;x \ne 2\\
TH3:B = 0\\
\to x = 4
\end{array}\)
c) \(Min = - \dfrac{1}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{{2x + 1 - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x + 1}} - \sqrt x } \right] + \dfrac{{2 - 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{2x + 1 - x + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {x - 2\sqrt x + 1} \right) + \dfrac{{2 - 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.{\left( {\sqrt x - 1} \right)^2}\\
= \sqrt x - 1 + \dfrac{{2 - 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - \sqrt x + 2 - 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x }}\\
b)B = A.\left( {\sqrt x - 1} \right)\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x }}.\left( {\sqrt x - 1} \right)\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}\left( {\sqrt x - 2} \right)}}{{\sqrt x }}\\
Do:x > 0;x \ne 1\\
\to \left\{ \begin{array}{l}
{\left( {\sqrt x - 1} \right)^2} > 0\\
\sqrt x > 0
\end{array} \right.\\
TH1:B > 0\\
\Leftrightarrow \sqrt x - 2 > 0\\
\to x > 4\\
TH2:B < 0\\
\to \sqrt x - 2 < 0\\
\to 0 < x < 4;x \ne 2\\
TH3:B = 0\\
\to \sqrt x - 2 = 0\\
\to x = 4\\
c)C = A:\sqrt x \\
= \dfrac{{x - 3\sqrt x + 2}}{{\sqrt x }}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{{x - 3\sqrt x + 2}}{x}\\
= 1 - \dfrac{3}{{\sqrt x }} + \dfrac{2}{x}\\
= \dfrac{2}{x} - 2.\sqrt {\dfrac{2}{x}} .\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8} - \dfrac{1}{8}\\
= {\left( {\sqrt {\dfrac{2}{x}} - \dfrac{3}{{2\sqrt 2 }}} \right)^2} - \dfrac{1}{8}\\
Do:{\left( {\sqrt {\dfrac{2}{x}} - \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x > 0\\
\to {\left( {\sqrt {\dfrac{2}{x}} - \dfrac{3}{{2\sqrt 2 }}} \right)^2} - \dfrac{1}{8} \ge - \dfrac{1}{8}\\
\to Min = - \dfrac{1}{8}\\
\Leftrightarrow \sqrt {\dfrac{2}{x}} - \dfrac{3}{{2\sqrt 2 }} = 0\\
\to \sqrt {\dfrac{2}{x}} = \dfrac{3}{{2\sqrt 2 }}\\
\to x = \dfrac{{16}}{9}
\end{array}\)
