$A =\dfrac{\sqrt x + 4}{\sqrt x -1}$
$B =\dfrac{3\sqrt x+1}{x + 2\sqrt x -3} - \dfrac{2}{\sqrt x +3}$
$(x\geqslant 0;\ x \ne 1)$
$+)\quad x = 9\Rightarrow \sqrt x = 3$
Thay vào $A$ ta được:
$\quad A =\dfrac{3 +4}{3 -1}=\dfrac72$
$+)\quad B = \dfrac{3\sqrt x +1}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}- \dfrac{2\left(\sqrt x - 1\right)}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}$
$\to B =\dfrac{3\sqrt x + 1 - 2\sqrt x + 2}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}$
$\to B = \dfrac{\sqrt x +3}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}$
$\to B =\dfrac{1}{\sqrt x -1}\quad$ (đpcm)
$+)\quad \dfrac{A}{B}\geqslant \dfrac x4 + 5$
$\Leftrightarrow \dfrac{\dfrac{\sqrt x +4}{\sqrt x -1}}{\dfrac{1}{\sqrt x -1}}\geqslant \dfrac x4 + 5$
$\Leftrightarrow \sqrt x + 4\geqslant \dfrac x4 + 5$
$\Leftrightarrow \dfrac x4 - \sqrt x + 1 \geqslant 0$
$\Leftrightarrow \left(\dfrac{\sqrt x}{2} - 1\right)^2 \geqslant 0$ (luôn đúng)
Vậy $x \geqslant 0;\ x\ne 1$
