a) ĐKXĐ : \(x>9;x\ge0\)
\(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
P =\(\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left[\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right]\)
P = \(\left[\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
P = \(\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
P = \(\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
P = \(\dfrac{-3}{\sqrt{x}+3}\)
Vậy với ĐKXĐ x>9 ; x \(\ge0\) thì biểu thức P rút gọn bằng \(\dfrac{-3}{\sqrt{x}+3}\)
b) ĐKXĐ : x>9 ; x≥0
Ta có x = \(4-2\sqrt{3}\)
\(\sqrt{x}=\sqrt{4-2\sqrt{3}}\)
\(\sqrt{x}=\sqrt{3-2\sqrt{3}+1}\)
\(\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(\sqrt{x}=\sqrt{3}-1\) ( thỏa mãn )
Thay \(\sqrt{x}=\sqrt{3}-1\) vào biểu thức P , ta được :
P = \(\dfrac{-3}{\sqrt{3}-1+3}\) = \(\dfrac{-3}{\sqrt{3}+2}\) = \(\dfrac{-3\left(\sqrt{3}-2\right)}{3-4}\) =\(\dfrac{-3\sqrt{3}+6}{-1}\) = \(3\sqrt{3}-6\)
Vậy với x = \(4-2\sqrt{3}\) thì P = \(3\sqrt{3}-6\)
