Question

Cho \(A=\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\); \(B=\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\) \(\left(x\ge0;x\ne9\right)\). Biết \(C=\dfrac{B}{A}\). Tìm \(x\in Z\) để \(C< \dfrac{-1}{3}\).

 

Answer 1

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

\(=\dfrac{-3}{2\sqrt{x}+4}\)

Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)

\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)

\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)

hay \(0\le x< \dfrac{25}{4}\)