Cho hai biểu thức $A=\dfrac{\sqrt[]{x}+4}{\sqrt[]{x}-1}$ và $B=\dfrac{3\sqrt[]{x}+1}{x+2\sqrt[]{x}-3}-\dfrac{2}{\sqrt[]{x}+3}$ (với $x\geq 0$, `xne 1

$A =\dfrac{\sqrt x + 4}{\sqrt x -1}$

$B =\dfrac{3\sqrt x+1}{x + 2\sqrt x -3} – \dfrac{2}{\sqrt x +3}$

$(x\geqslant 0;\ x \ne 1)$

$+)\quad x = 9\Rightarrow \sqrt x = 3$

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 $x \geqslant 0;\ x\ne 1$