Giải thích các bước giải:
a. khi x = 64
A = $\frac{\sqrt{64}+2}{\sqrt{64}-3}$ = $\frac{8+2}{8-3}$ = $\frac{10}{5} = 2$$=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{(\sqrt{x}-3)(\sqrt{x}+3) }$ $\frac{-3\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3) }$
b. ĐK: $x\geq 0, x$ $\neq 9$
$B =\frac{2\sqrt{x}}{\sqrt{x}+3} + \frac{\sqrt{x}}{\sqrt{x}-3} - \frac{3x+3}{x-9}$
$= \frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-3x-3}{(\sqrt{x}-3) (\sqrt{x}+3) }$
c.
$P = \frac{B}{A} = \frac{\frac{-3\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}}{\frac{\sqrt{x}+2}{\sqrt{x}-3}} = \frac{-3\sqrt{x}-3}{(\sqrt{x}+2)(\sqrt{x}+3)}$ \\ $P<\frac{1}{3}⇔-9\sqrt{x}-9 < x + 5\sqrt{x}+6⇔x+14\sqrt{x}+15 >0 ⇔ ...$
