A=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
ĐKXĐ :x\(\ne\)9,x\(\ge\)0
<=> \(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}\)
=\(\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
=\(\dfrac{3\sqrt{x}-9}{x-9}\)=\(\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
ta có : A=1/3 => \(\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}=>\sqrt{x}+3=9\)
=> x=36
vậy giá trị của x=36 khi A=1/3
B=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-1}\right):\dfrac{1}{\sqrt{x}+1}\)
ĐKXĐ: x\(\ne\)1 ,x\(\ge\)0
<=> \(\dfrac{\sqrt{x}+1-\sqrt{x}}{x-1}:\dfrac{1}{\sqrt{x}+1}\)
=\(\dfrac{1}{x-1}:\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x-1}=\dfrac{1}{\sqrt{x}-1}\)
ta có : B<0 =>\(\dfrac{1}{\sqrt{x}-1}< 0\)
=> 1< \(\sqrt{x}-1\)=> \(\sqrt{x}\)>2=>x>4
vậy x>4 khi B<0
\(ĐKXĐ:x\ne9\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9} =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{3x+9}{x-9}=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)
