Question

Cho biểu thức :

P = ( \(\dfrac{x}{\sqrt{x}+3}-\dfrac{x+1}{\sqrt{x}-3}+\dfrac{6x+\sqrt{x}}{x-9}\) ) : ( \(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-1\) ) Với x > 0 ; x khác 9

a) Rút gọn P

b) Tính P khi x = \(12+6\sqrt{3}\)

Answer 1

\(P=\left(\dfrac{x}{\sqrt{x}+3}-\dfrac{x+1}{\sqrt{x}-3}+\dfrac{6x+\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-1\right)\)

\(P=\dfrac{x\left(\sqrt{x}-3\right)-\left(x+1\right)\left(\sqrt{x}+3\right)+6x+\sqrt{x}}{x-9}:\dfrac{\left(\sqrt{x}-3\right)^2-x+9}{x-9}\)

\(P=\dfrac{x\sqrt{x}-3x-x\sqrt{x}-3x-\sqrt{x}-3+6x+\sqrt{x}}{x-9}:\dfrac{x-6\sqrt{x}+9-x+9}{x-9}\)

\(P=-\dfrac{3}{x-9}:\dfrac{-6\sqrt{x}+18}{x-9}=-\dfrac{3}{x-9}.\dfrac{x-9}{-6\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{1}{2\sqrt{x}-6}\)

b. \(x=12+6\sqrt{3}=9+2.3.\sqrt{3}+3=\left(3+\sqrt{3}\right)^2\Rightarrow\sqrt{x}=3+\sqrt{3}\)

\(P=\dfrac{1}{2.3+2\sqrt{3}-6}=\dfrac{1}{2\sqrt{3}}=\dfrac{\sqrt{3}}{6}\)