$1. P=\left (\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x-3}{x-9} \right ) : \left (\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1 \right )$ $( x \geq 0 ; x \neq 9 )$
$=\left [\dfrac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{3x-3}{(\sqrt{x}-3)(\sqrt{x}+3)} \right ] : \dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}$
$=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{(\sqrt{x}-3)(\sqrt{x}+3)} . \dfrac{\sqrt{x}-3}{\sqrt{x}+1}$
$=\dfrac{-3\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}+1)}$
Vậy $P=\dfrac{-3\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}+1)}$
$2. x = \dfrac{4-\sqrt{7}}{2}=\dfrac{8-2\sqrt{7}}{4}=\dfrac{(\sqrt{7}-1)^2}{2^2} ( t/m )$
$⇒P=\dfrac{-3\sqrt{\dfrac{(\sqrt{7}-1)^2}{2^2}}+3}{\left (\sqrt{\dfrac{(\sqrt{7}-1)^2}{2^2}}+3 \right ).\left (\sqrt{\dfrac{(\sqrt{7}-1)^2}{2^2}}+1 \right )}$
$=\dfrac{\dfrac{-3(\sqrt{7}-1)}{2}+3}{\left (\dfrac{\sqrt{7}-1}{2}+3 \right ).\left (\dfrac{\sqrt{7}-1}{2}+1 \right )}$
$=\dfrac{\dfrac{-3\sqrt{7}+3+6}{2}}{\left (\dfrac{\sqrt{7}-1+6}{2} \right ).\left (\dfrac{\sqrt{7}-1+2}{2} \right )}$
$=\dfrac{9-3\sqrt{7}}{2} : \dfrac{(\sqrt{7}+5)(\sqrt{7}+1)}{4}$
$=\dfrac{3(3-\sqrt{7})}{2} . \dfrac{4}{7+6\sqrt{7}+5}$
$=\dfrac{6(3-\sqrt{7})}{12+6\sqrt{7}}=\dfrac{6(3-\sqrt{7})}{6(2+\sqrt{7})}$
$=\dfrac{3-\sqrt{7}}{2+\sqrt{7}}=\dfrac{(3-\sqrt{7})(2-\sqrt{7})}{(2+\sqrt{7})(2-\sqrt{7})}$
$=\dfrac{6-3\sqrt{7}-2\sqrt{7}+7}{4-7}$
$=\dfrac{13-5\sqrt{7}}{-3}$
$=\dfrac{5\sqrt{7}-13}{3}$
Vậy $P=\dfrac{5\sqrt{7}-13}{3}$ với $x=\dfrac{4-\sqrt{7}}{2}$
