Đáp án:x>9
Giải thích các bước giải:
$B = \left( {\frac{{\sqrt x }}{{x - 4}} + \frac{1}{{\sqrt x - 2}} + \frac{{2\sqrt x }}{{\sqrt x + 2}}} \right):\left( {\frac{{x\sqrt x + 1}}{{x\sqrt x - 4\sqrt x }}} \right)$
ĐK: $\{ _{x > 0}^{x \ne 4}$
$\begin{array}{l}
B = \left( {\frac{{\sqrt x }}{{x - 4}} + \frac{1}{{\sqrt x - 2}} + \frac{{2\sqrt x }}{{\sqrt x + 2}}} \right):\left( {\frac{{x\sqrt x + 1}}{{x\sqrt x - 4\sqrt x }}} \right)\\
= \left( {\frac{{\sqrt x + \sqrt x + 2 + 2\sqrt x (\sqrt x - 2)}}{{(\sqrt x - 2)(\sqrt x + 2)}}} \right):\frac{{{{(\sqrt x )}^3} + 1}}{{\sqrt x (x - 4)}}\\
= \left( {\frac{{2x - 2\sqrt x + 2}}{{x - 4}}} \right).\frac{{\sqrt x (x - 4)}}{{(\sqrt x + 1)(x - \sqrt x + 1)}}\\
= \frac{{2(x - \sqrt x + 1)}}{{x - 4}}.\frac{{\sqrt x (x - 4)}}{{(\sqrt x + 1)(x - \sqrt x + 1)}}\\
= \frac{{2\sqrt x }}{{\sqrt x + 1}}
\end{array}$
Để
$\begin{array}{l}
2{B^2} > 3B\\
< = > 2{\left( {\frac{{2\sqrt x }}{{\sqrt x + 1}}} \right)^2} > 3.\frac{{2\sqrt x }}{{\sqrt x + 1}}\\
< = > \frac{{8x}}{{{{(\sqrt x + 1)}^2}}} > \frac{{6\sqrt x }}{{\sqrt x + 1}}\\
< = > \frac{{8x - 6\sqrt x (\sqrt x + 1)}}{{{{(\sqrt x + 1)}^2}}} > 0\\
< = > \frac{{2x - 6\sqrt x }}{{{{(\sqrt x + 1)}^2}}} > 0\\
= > 2x - 6\sqrt x > 0\\
< = > 2\sqrt x (\sqrt x - 3) > 0\\
< = > \{ _{\sqrt x - 3 > 0}^{2\sqrt x > 0}\\
< = > \{ _{x > 9}^{x > 0} = > x > 9
\end{array}$
