Bài 2:
a: \(B=\dfrac{\sqrt{x}-3+x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b: Để B>1/2 thì B-1/2>0
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}-\dfrac{1}{2}>0\)
\(\Leftrightarrow2\sqrt{x}+2-\sqrt{x}-3>0\)
=>x>1
c: Để \(B=\dfrac{3}{2}:\sqrt{x}=\dfrac{3}{2\sqrt{x}}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{3}{2\sqrt{x}}\)
\(\Leftrightarrow2x+2\sqrt{x}-3\sqrt{x}-9=0\)
\(\Leftrightarrow2x-1\sqrt{x}-9=0\)
Đặt \(\sqrt{x}=a\)
Pt sẽ là \(2a^2-a-9=0\)
\(\Leftrightarrow a=\dfrac{\sqrt{73}+1}{4}\)
hay \(x=\dfrac{\left(\sqrt{73}+1\right)^2}{16}\)
