Đáp án:
\(\begin{array}{l}
a)B = \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
b)Min = 2\sqrt 3 + 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \left[ {\dfrac{{x\sqrt x + 3x - 3x - 1}}{{x + 3\sqrt x }}} \right].\dfrac{{\sqrt x + 3}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{\sqrt x - 1}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
b)P = A.B = \dfrac{{\sqrt x }}{{\sqrt x - 1}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{x - 2\sqrt x + 1 + 3\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2} + 3\sqrt x }}{{\sqrt x - 1}}\\
= \left( {\sqrt x - 1} \right) + \dfrac{{3\sqrt x }}{{\sqrt x - 1}}\\
= \left( {\sqrt x - 1} \right) + \dfrac{{3\left( {\sqrt x - 1} \right) + 3}}{{\sqrt x - 1}}\\
= \left( {\sqrt x - 1} \right) + 3 + \dfrac{3}{{\sqrt x - 1}}\\
Xét:1 < x \\
BDT:Co - si:\left( {\sqrt x - 1} \right) + \dfrac{3}{{\sqrt x - 1}} \ge 2\sqrt {\left( {\sqrt x - 1} \right).\dfrac{3}{{\sqrt x - 1}}} = 2\sqrt 3 \\
\to \left( {\sqrt x - 1} \right) + \dfrac{3}{{\sqrt x - 1}} + 3 \ge 2\sqrt 3 + 3\\
\to Min = 2\sqrt 3 + 3\\
\Leftrightarrow \left( {\sqrt x - 1} \right) = \dfrac{3}{{\sqrt x - 1}}\\
\Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 3\\
\to \sqrt x - 1 = \sqrt 3 \\
\to \sqrt x = 1 + \sqrt 3 \\
\to x = 4 + 2\sqrt 3
\end{array}\)
\(\begin{array}{l}
Xét:0 < x < 1\\
P = \left( {\sqrt x - 1} \right) + 3 + \dfrac{3}{{\sqrt x - 1}}\\
= \sqrt x + 2 + \dfrac{3}{{\sqrt x - 1}}\\
P\min \Leftrightarrow \dfrac{3}{{\sqrt x - 1}}\min \\
\Leftrightarrow \left( {\sqrt x - 1} \right)\max \\
\Leftrightarrow \sqrt x \max
\end{array}\)
⇒ Không tồn tại x trong khoảng 0<x<1 để P đạt GTNN
