Đáp án:
b. \(\dfrac{{3x + 3\sqrt x }}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:x = \dfrac{{16}}{9}\\
\to A = \dfrac{3}{{3\sqrt {\dfrac{{16}}{9}} + 1}} = \dfrac{3}{{3.\dfrac{4}{3} + 1}} = \dfrac{3}{5}\\
b.DK:x \ge 0;x \ne \dfrac{1}{9}\\
B = \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right) - 3\sqrt x + 1 + 8\sqrt x }}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}\\
= \dfrac{{3x - 2\sqrt x - 1 + 5\sqrt x + 1}}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}\\
= \dfrac{{3x + 3\sqrt x }}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}\\
c.DK:x > 0\\
P = A:B = \dfrac{3}{{3\sqrt x + 1}}:\dfrac{{3x + 3\sqrt x }}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}\\
= \dfrac{3}{{3\sqrt x + 1}}.\dfrac{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}{{3\left( {x + \sqrt x } \right)}}\\
= \dfrac{{3\sqrt x - 1}}{{x + \sqrt x }}\\
P < \dfrac{5}{9}\\
\to \dfrac{{3\sqrt x - 1}}{{x + \sqrt x }} < \dfrac{5}{9}\\
\to \dfrac{{27\sqrt x - 9 - 5x - 5\sqrt x }}{{9\sqrt x \left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{ - 5x + 22\sqrt x - 9}}{{9\sqrt x \left( {\sqrt x + 1} \right)}} < 0\\
\to - 5x + 22\sqrt x - 9 < 0\\
\left( {Do:\left\{ \begin{array}{l}
\sqrt x > 0\\
\sqrt x + 1 > 0
\end{array} \right.\forall x > 0} \right)\\
\to \dfrac{{11 - 2\sqrt {19} }}{5} < \sqrt x < \dfrac{{11 + 2\sqrt {19} }}{5}\\
\to {\left( {\dfrac{{11 - 2\sqrt {19} }}{5}} \right)^2} < x < {\left( {\dfrac{{11 + 2\sqrt {19} }}{5}} \right)^2}
\end{array}\)
