a) P = ($\frac{\sqrt{x}}{\sqrt{x}+3}$ + $\frac{2\sqrt{x}}{\sqrt{x}-3}$ - $\frac{3x+3}{x-9}$) : ($\frac{2\sqrt{x}+2}{\sqrt{x}+3}$ -1)
P = ($\frac{\sqrt{x}}{\sqrt{x}+3}$ + $\frac{2\sqrt{x}}{\sqrt{x}-3}$ - $\frac{3x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}$) : ($\frac{2\sqrt{x}+2}{\sqrt{x}+3}$ - $\frac{\sqrt{x}+3}{\sqrt{x}+3}$ )
P = $\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)-(3x+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}$ : $\frac{2\sqrt{x}+2-(\sqrt{x}+3)}{\sqrt{x}+3}$
P = $\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-3}{(\sqrt{x}+3)(\sqrt{x}-3)}$ : $\frac{2\sqrt{x}+2-\sqrt{x}-3}{\sqrt{x}+3}$
P = $\frac{3\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}$ : $\frac{\sqrt{x}-1}{\sqrt{x}+3}$
P = $\frac{3\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}$ . $\frac{\sqrt{x}+3}{\sqrt{x}-1}$
P = $\frac{3(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-3)}$ . $\frac{\sqrt{x}+3}{\sqrt{x}-1}$
P = $\frac{3}{\sqrt{x}-3}$
b) Để P<1 thì $\frac{3}{\sqrt{x}-3}$ < 1
<=>$\frac{3}{\sqrt{x}-3}$ < $\frac{\sqrt{x}-3}{\sqrt{x}-3}$
<=> 3 < $\sqrt{x}$ - 3
<=> 3 + 3 < $\sqrt{x}$
<=> $\sqrt{x}$ > 6
<=> x > 36
Vậy để P<1 thì x >36
