Question

Cho \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\) với \(x\ge0;x\ne9\)

a. Rút gọn A

b. Tìm x để \(A>\dfrac{1}{3}\)

Answer 1

a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)

\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)

\(=\dfrac{2}{\sqrt{x}+3}\)

b: Để \(A>\dfrac{1}{3}\) thì \(A-\dfrac{1}{3}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)

Answer 2

a) \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne0\right)\)

\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\)

b) \(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\)

\(\Leftrightarrow6>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)