Question

Answer 1

1, Với x = 1

\(A=\dfrac{1+2}{9}=\dfrac{1}{3}\)

2, Với x >= 0 ; x khác 4;9

\(B=\left(\dfrac{3\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{x-9}{\sqrt{x}-3}\)

\(=\left(\dfrac{3+\sqrt{x}}{\sqrt{x}+2}\right):\left(\sqrt{x}+3\right)=\dfrac{1}{\sqrt{x}+2}\)

Vậy ta có đpcm 

3, Ta có \(A+B=\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{9}=\dfrac{2}{3}\)

\(\Leftrightarrow9+x+4\sqrt{x}+4=6\left(\sqrt{x}+2\right)\)

\(\Leftrightarrow x-2\sqrt{x}+1=0\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\left(tm\right)\)