Bài 2:
a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(B=\left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right):\dfrac{3}{x-3}\)
\(=\dfrac{x+3-x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{3}\)
\(=\dfrac{2}{x+3}\)
b: Để \(B>\dfrac{1}{3}\) thì \(B-\dfrac{1}{3}>0\)
\(\Leftrightarrow\dfrac{2}{x+3}-\dfrac{1}{3}>0\)
\(\Leftrightarrow\dfrac{6-x-3}{3\left(x+3\right)}>0\)
\(\Leftrightarrow\dfrac{3-x}{3\left(x+3\right)}>0\)
\(\Leftrightarrow\dfrac{1}{3}\cdot\dfrac{x-3}{x+3}< 0\)
\(\Leftrightarrow-3< x< 3\)
Đúng 0
Bình luận (0)
