Đáp án:
3) Min=2
Giải thích các bước giải:
\(\begin{array}{l}
1){x^2} - 3x + 2 = 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1\left( l \right)
\end{array} \right.\\
Thay:x = 2\\
\to A = \dfrac{2}{{\sqrt 2 - 1}} = \dfrac{{2\left( {\sqrt 2 + 1} \right)}}{{2 - 1}} = 2\sqrt 2 + 2\\
2)B = \left[ {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]:\left[ {\dfrac{{2\sqrt x - 2 + x + 2}}{{x\left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{x\left( {\sqrt x - 1} \right)}}{{x + 2\sqrt x }}\\
= \dfrac{x}{{\sqrt x + 1}}\\
\sqrt P \le \dfrac{2}{3}\\
\to P \le \dfrac{4}{9}\\
\to \dfrac{x}{{\sqrt x + 1}} \le \dfrac{4}{9}\\
\to \dfrac{{9x - 4\sqrt x - 4}}{{9\left( {\sqrt x + 1} \right)}} \le 0\\
\to 9x - 4\sqrt x - 4 \le 0\left( {do:\sqrt x + 1 > 0\forall x > 0} \right)\\
\to \dfrac{{2 - 2\sqrt {10} }}{9} \le \sqrt x \le \dfrac{{2 + 2\sqrt {10} }}{9}\\
\to 0 \le x \le \dfrac{{44 + 8\sqrt {10} }}{{81}};x \ne 1\\
3)\sqrt A = \sqrt {\dfrac{x}{{\sqrt x - 1}}} = \sqrt {\dfrac{{x - 1 + 1}}{{\sqrt x - 1}}} \\
= \sqrt {\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1}}{{\sqrt x - 1}}} \\
= \sqrt {\left( {\sqrt x + 1} \right) + \dfrac{1}{{\sqrt x - 1}}} \\
= \sqrt {\left( {\sqrt x - 1} \right) + \dfrac{1}{{\sqrt x - 1}} + 2} \\
Do:x > 0\\
BDT:Co - si:\\
\left( {\sqrt x - 1} \right) + \dfrac{1}{{\sqrt x - 1}} \ge 2\sqrt {\left( {\sqrt x - 1} \right).\dfrac{1}{{\sqrt x - 1}}} = 2\\
\to \left( {\sqrt x - 1} \right) + \dfrac{1}{{\sqrt x - 1}} + 2 \ge 4\\
\to \sqrt {\left( {\sqrt x - 1} \right) + \dfrac{1}{{\sqrt x - 1}} + 2} \ge 2\\
\to Min = 2\\
\Leftrightarrow \left( {\sqrt x - 1} \right) = \dfrac{1}{{\sqrt x - 1}}\\
\to {\left( {\sqrt x - 1} \right)^2} = 1\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\left( {TM} \right)\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
